How to find the roots of $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$?How to integrate $sqrt1-sin 2x$?Show that $5cos^2x - 2sqrt3sin xcos x + 3sin^2x = cos 2x - sqrt3sin2x + 4$How to Simplify $-frac14sinfrac x4 + fracsqrt34cosfrac x4 = 0$?Number of solutions of $8sin(x)=fracsqrt3cos(x)+frac1sin(x)$Solve $2cos^2x=sqrt3sin2x$.If $fracsin xsin y= 3$ and $fraccos xcos y= frac12$, then find $fracsin2xsin2y+fraccos2xcos2y$Find the general value of $x$ satisfying $sin x=frac12$ and $cos x=-fracsqrt 32$How to solve $sin^2(x)+sin2x+2cos^2(x)=0$Area bounded by $y=sqrtfrac1+sinxcosx$ and $y=sqrtfrac1-sinxcosx$Interval in which $(cos p-1)x^2+cos p.x+sin p=0$ has real roots

How does airport security verify that you can carry a battery bank over 100 Wh?

Should I tell my boss the work he did was worthless

How to create a hard link to an inode (ext4)?

Could a cubesat be propelled to the moon?

Why the color red for the Republican Party

Time travel short story where dinosaur doesn't taste like chicken

Am I not good enough for you?

Can you reject a postdoc offer after the PI has paid a large sum for flights/accommodation for your visit?

Offered promotion but I'm leaving. Should I tell?

Do I really need to have a scientific explanation for my premise?

Best approach to update all entries in a list that is paginated?

Examples of a statistic that is not independent of sample's distribution?

Grey hair or white hair

They call me Inspector Morse

Virginia employer terminated employee and wants signing bonus returned

Are babies of evil humanoid species inherently evil?

PTIJ: How can I halachically kill a vampire?

How strictly should I take "Candidates must be local"?

What is the chance of making a successful appeal to dismissal decision from a PhD program after failing the qualifying exam in the 2nd attempt?

How much stiffer are 23c tires over 28c?

How do I locate a classical quotation?

Built-In Shelves/Bookcases - IKEA vs Built

My story is written in English, but is set in my home country. What language should I use for the dialogue?

Why is there a voltage between the mains ground and my radiator?



How to find the roots of $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$?


How to integrate $sqrt1-sin 2x$?Show that $5cos^2x - 2sqrt3sin xcos x + 3sin^2x = cos 2x - sqrt3sin2x + 4$How to Simplify $-frac14sinfrac x4 + fracsqrt34cosfrac x4 = 0$?Number of solutions of $8sin(x)=fracsqrt3cos(x)+frac1sin(x)$Solve $2cos^2x=sqrt3sin2x$.If $fracsin xsin y= 3$ and $fraccos xcos y= frac12$, then find $fracsin2xsin2y+fraccos2xcos2y$Find the general value of $x$ satisfying $sin x=frac12$ and $cos x=-fracsqrt 32$How to solve $sin^2(x)+sin2x+2cos^2(x)=0$Area bounded by $y=sqrtfrac1+sinxcosx$ and $y=sqrtfrac1-sinxcosx$Interval in which $(cos p-1)x^2+cos p.x+sin p=0$ has real roots













3












$begingroup$


Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.



The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All



I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?



My attempt:



$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$



or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$



or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$



or, $sin(x+pi/12)=sin2x$



or, $x=pi/12$










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
    $endgroup$
    – MrAP
    2 days ago







  • 1




    $begingroup$
    Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    @MinusOne-Twelfth, Check my question now.
    $endgroup$
    – MrAP
    2 days ago






  • 1




    $begingroup$
    You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
    $endgroup$
    – Minus One-Twelfth
    2 days ago
















3












$begingroup$


Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.



The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All



I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?



My attempt:



$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$



or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$



or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$



or, $sin(x+pi/12)=sin2x$



or, $x=pi/12$










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
    $endgroup$
    – MrAP
    2 days ago







  • 1




    $begingroup$
    Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    @MinusOne-Twelfth, Check my question now.
    $endgroup$
    – MrAP
    2 days ago






  • 1




    $begingroup$
    You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
    $endgroup$
    – Minus One-Twelfth
    2 days ago














3












3








3


1



$begingroup$


Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.



The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All



I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?



My attempt:



$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$



or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$



or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$



or, $sin(x+pi/12)=sin2x$



or, $x=pi/12$










share|cite|improve this question











$endgroup$




Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.



The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All



I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?



My attempt:



$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$



or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$



or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$



or, $sin(x+pi/12)=sin2x$



or, $x=pi/12$







trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







MrAP

















asked 2 days ago









MrAPMrAP

1,16721432




1,16721432







  • 1




    $begingroup$
    How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
    $endgroup$
    – MrAP
    2 days ago







  • 1




    $begingroup$
    Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    @MinusOne-Twelfth, Check my question now.
    $endgroup$
    – MrAP
    2 days ago






  • 1




    $begingroup$
    You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
    $endgroup$
    – Minus One-Twelfth
    2 days ago













  • 1




    $begingroup$
    How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
    $endgroup$
    – MrAP
    2 days ago







  • 1




    $begingroup$
    Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    @MinusOne-Twelfth, Check my question now.
    $endgroup$
    – MrAP
    2 days ago






  • 1




    $begingroup$
    You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
    $endgroup$
    – Minus One-Twelfth
    2 days ago








1




1




$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago





$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago













$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago





$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago





1




1




$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago





$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago













$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago




$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago




1




1




$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago





$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago











2 Answers
2






active

oldest

votes


















5












$begingroup$

Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$




Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$



enter image description here






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
    $endgroup$
    – Matteo
    2 days ago







  • 1




    $begingroup$
    Oh. I got it. Mine generalizes both odd and even n in one equation.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
    $endgroup$
    – Matteo
    2 days ago



















2












$begingroup$

Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$



Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I had got that equation at the end. Then I equated the angles to get $x=15^circ$
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP I got also $55^circ.$
    $endgroup$
    – Michael Rozenberg
    2 days ago










  • $begingroup$
    How did you get that value?
    $endgroup$
    – MrAP
    2 days ago










  • $begingroup$
    @MrAP I added something. See now.
    $endgroup$
    – Michael Rozenberg
    2 days ago










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142283%2fhow-to-find-the-roots-of-frac-sqrt3-1-sin-x-frac-sqrt31-cos-x-4%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$




Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$



enter image description here






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
    $endgroup$
    – Matteo
    2 days ago







  • 1




    $begingroup$
    Oh. I got it. Mine generalizes both odd and even n in one equation.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
    $endgroup$
    – Matteo
    2 days ago
















5












$begingroup$

Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$




Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$



enter image description here






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
    $endgroup$
    – Matteo
    2 days ago







  • 1




    $begingroup$
    Oh. I got it. Mine generalizes both odd and even n in one equation.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
    $endgroup$
    – Matteo
    2 days ago














5












5








5





$begingroup$

Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$




Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$



enter image description here






share|cite|improve this answer











$endgroup$



Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$




Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









MatteoMatteo

1,032313




1,032313











  • $begingroup$
    How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
    $endgroup$
    – Matteo
    2 days ago







  • 1




    $begingroup$
    Oh. I got it. Mine generalizes both odd and even n in one equation.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
    $endgroup$
    – Matteo
    2 days ago

















  • $begingroup$
    How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
    $endgroup$
    – Matteo
    2 days ago







  • 1




    $begingroup$
    Oh. I got it. Mine generalizes both odd and even n in one equation.
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
    $endgroup$
    – Matteo
    2 days ago
















$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago





$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago













$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago





$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago





1




1




$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago





$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago













$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago





$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago












2












$begingroup$

Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$



Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I had got that equation at the end. Then I equated the angles to get $x=15^circ$
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP I got also $55^circ.$
    $endgroup$
    – Michael Rozenberg
    2 days ago










  • $begingroup$
    How did you get that value?
    $endgroup$
    – MrAP
    2 days ago










  • $begingroup$
    @MrAP I added something. See now.
    $endgroup$
    – Michael Rozenberg
    2 days ago















2












$begingroup$

Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$



Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I had got that equation at the end. Then I equated the angles to get $x=15^circ$
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP I got also $55^circ.$
    $endgroup$
    – Michael Rozenberg
    2 days ago










  • $begingroup$
    How did you get that value?
    $endgroup$
    – MrAP
    2 days ago










  • $begingroup$
    @MrAP I added something. See now.
    $endgroup$
    – Michael Rozenberg
    2 days ago













2












2








2





$begingroup$

Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$



Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?






share|cite|improve this answer











$endgroup$



Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$



Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Michael RozenbergMichael Rozenberg

108k1895200




108k1895200











  • $begingroup$
    I had got that equation at the end. Then I equated the angles to get $x=15^circ$
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP I got also $55^circ.$
    $endgroup$
    – Michael Rozenberg
    2 days ago










  • $begingroup$
    How did you get that value?
    $endgroup$
    – MrAP
    2 days ago










  • $begingroup$
    @MrAP I added something. See now.
    $endgroup$
    – Michael Rozenberg
    2 days ago
















  • $begingroup$
    I had got that equation at the end. Then I equated the angles to get $x=15^circ$
    $endgroup$
    – MrAP
    2 days ago











  • $begingroup$
    @MrAP I got also $55^circ.$
    $endgroup$
    – Michael Rozenberg
    2 days ago










  • $begingroup$
    How did you get that value?
    $endgroup$
    – MrAP
    2 days ago










  • $begingroup$
    @MrAP I added something. See now.
    $endgroup$
    – Michael Rozenberg
    2 days ago















$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago





$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago













$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago




$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago












$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago




$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago












$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago




$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago

















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142283%2fhow-to-find-the-roots-of-frac-sqrt3-1-sin-x-frac-sqrt31-cos-x-4%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye