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
$begingroup$
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, $x=pi/12$
trigonometry
$endgroup$
add a comment |
$begingroup$
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, $x=pi/12$
trigonometry
$endgroup$
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
add a comment |
$begingroup$
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, $x=pi/12$
trigonometry
$endgroup$
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
trigonometry
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$X^2+Y^2 = 1.$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
$endgroup$
$begingroup$
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
add a comment |
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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
$$X^2+Y^2 = 1.$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$X^2+Y^2 = 1.$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$X^2+Y^2 = 1.$$
$endgroup$
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
$$X^2+Y^2 = 1.$$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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?
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
$endgroup$
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?
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
add a comment |
$begingroup$
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
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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