Solve tan(x)+cos(x)=1/2 The Next CEO of Stack OverflowEvaluate the $sin$, $cos$ and $tan$ without using calculator?Solve trigonometric equation $tantheta + sectheta =2cos theta$finding variables of one side of the equation given cos and tanTrying to solve $sqrt7-4sqrt2 sin x=2cos(x)-sqrt2 tan(x)$In $triangle ABC$, if $cos Acos Bcos C=frac13$, then $tan Atan B+tan B tan C+tan Ctan A =text???$Verify $fraccot x -tan xcos x + sin x=fraccos x - sin xsin x cos x$Given $tantheta=7/9$ and $costheta<0$, find $tan(theta/2)$.What is $cos[2tan ^-1(x^2)]$Given $cos x = -frac14 sqrt3$, find possible values of $tan x$.Find $sin x$ if $cos x=tan y$, $cos y=tan z$, $cos z=tan x$
What does "shotgun unity" refer to here in this sentence?
Man transported from Alternate World into ours by a Neutrino Detector
Expressing the idea of having a very busy time
Cannot shrink btrfs filesystem although there is still data and metadata space left : ERROR: unable to resize '/home': No space left on device
Can I calculate next year's exemptions based on this year's refund/amount owed?
Is French Guiana a (hard) EU border?
Are the names of these months realistic?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
What day is it again?
how one can write a nice vector parser, something that does pgfvecparseA=B-C; D=E x F;
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
From jafe to El-Guest
Is it ever safe to open a suspicious HTML file (e.g. email attachment)?
Does higher Oxidation/ reduction potential translate to higher energy storage in battery?
How did Beeri the Hittite come up with naming his daughter Yehudit?
Does Germany produce more waste than the US?
Do scriptures give a method to recognize a truly self-realized person/jivanmukta?
Reference request: Grassmannian and Plucker coordinates in type B, C, D
Players Circumventing the limitations of Wish
What would be the main consequences for a country leaving the WTO?
What is the difference between "hamstring tendon" and "common hamstring tendon"?
Computationally populating tables with probability data
Strange use of "whether ... than ..." in official text
How to find image of a complex function with given constraints?
Solve tan(x)+cos(x)=1/2
The Next CEO of Stack OverflowEvaluate the $sin$, $cos$ and $tan$ without using calculator?Solve trigonometric equation $tantheta + sectheta =2cos theta$finding variables of one side of the equation given cos and tanTrying to solve $sqrt7-4sqrt2 sin x=2cos(x)-sqrt2 tan(x)$In $triangle ABC$, if $cos Acos Bcos C=frac13$, then $tan Atan B+tan B tan C+tan Ctan A =text???$Verify $fraccot x -tan xcos x + sin x=fraccos x - sin xsin x cos x$Given $tantheta=7/9$ and $costheta<0$, find $tan(theta/2)$.What is $cos[2tan ^-1(x^2)]$Given $cos x = -frac14 sqrt3$, find possible values of $tan x$.Find $sin x$ if $cos x=tan y$, $cos y=tan z$, $cos z=tan x$
$begingroup$
Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
fracsqrt1-c^2c+c=frac12
$$
$$
sqrt1-c^2+c^2=frac12c
$$
$$
1-c^2=c^2(frac12-c)^2=c^2(frac14-c+c^2)
$$
$$
c^4-c^3+frac54c^2-1=0
$$
trigonometry
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
$endgroup$
add a comment |
$begingroup$
Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
fracsqrt1-c^2c+c=frac12
$$
$$
sqrt1-c^2+c^2=frac12c
$$
$$
1-c^2=c^2(frac12-c)^2=c^2(frac14-c+c^2)
$$
$$
c^4-c^3+frac54c^2-1=0
$$
trigonometry
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
$endgroup$
$begingroup$
Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2
$endgroup$
– Dr. Mathva
Mar 19 at 21:08
$begingroup$
Of course it is possible (cf. Ferrari's formula for the quartic, for example).
$endgroup$
– Allawonder
Mar 19 at 21:19
$begingroup$
@Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics
$endgroup$
– pasaba por aqui
Mar 19 at 21:23
$begingroup$
@pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was.
$endgroup$
– Allawonder
Mar 19 at 21:26
$begingroup$
@Allawonder: yes, your comment is correct, I only adding context
$endgroup$
– pasaba por aqui
Mar 19 at 21:29
add a comment |
$begingroup$
Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
fracsqrt1-c^2c+c=frac12
$$
$$
sqrt1-c^2+c^2=frac12c
$$
$$
1-c^2=c^2(frac12-c)^2=c^2(frac14-c+c^2)
$$
$$
c^4-c^3+frac54c^2-1=0
$$
trigonometry
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
$endgroup$
Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
fracsqrt1-c^2c+c=frac12
$$
$$
sqrt1-c^2+c^2=frac12c
$$
$$
1-c^2=c^2(frac12-c)^2=c^2(frac14-c+c^2)
$$
$$
c^4-c^3+frac54c^2-1=0
$$
trigonometry
trigonometry
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
asked Mar 19 at 21:04
pasaba por aquipasaba por aqui
454316
454316
$begingroup$
Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2
$endgroup$
– Dr. Mathva
Mar 19 at 21:08
$begingroup$
Of course it is possible (cf. Ferrari's formula for the quartic, for example).
$endgroup$
– Allawonder
Mar 19 at 21:19
$begingroup$
@Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics
$endgroup$
– pasaba por aqui
Mar 19 at 21:23
$begingroup$
@pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was.
$endgroup$
– Allawonder
Mar 19 at 21:26
$begingroup$
@Allawonder: yes, your comment is correct, I only adding context
$endgroup$
– pasaba por aqui
Mar 19 at 21:29
add a comment |
$begingroup$
Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2
$endgroup$
– Dr. Mathva
Mar 19 at 21:08
$begingroup$
Of course it is possible (cf. Ferrari's formula for the quartic, for example).
$endgroup$
– Allawonder
Mar 19 at 21:19
$begingroup$
@Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics
$endgroup$
– pasaba por aqui
Mar 19 at 21:23
$begingroup$
@pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was.
$endgroup$
– Allawonder
Mar 19 at 21:26
$begingroup$
@Allawonder: yes, your comment is correct, I only adding context
$endgroup$
– pasaba por aqui
Mar 19 at 21:29
$begingroup$
Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2
$endgroup$
– Dr. Mathva
Mar 19 at 21:08
$begingroup$
Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2
$endgroup$
– Dr. Mathva
Mar 19 at 21:08
$begingroup$
Of course it is possible (cf. Ferrari's formula for the quartic, for example).
$endgroup$
– Allawonder
Mar 19 at 21:19
$begingroup$
Of course it is possible (cf. Ferrari's formula for the quartic, for example).
$endgroup$
– Allawonder
Mar 19 at 21:19
$begingroup$
@Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics
$endgroup$
– pasaba por aqui
Mar 19 at 21:23
$begingroup$
@Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics
$endgroup$
– pasaba por aqui
Mar 19 at 21:23
$begingroup$
@pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was.
$endgroup$
– Allawonder
Mar 19 at 21:26
$begingroup$
@pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was.
$endgroup$
– Allawonder
Mar 19 at 21:26
$begingroup$
@Allawonder: yes, your comment is correct, I only adding context
$endgroup$
– pasaba por aqui
Mar 19 at 21:29
$begingroup$
@Allawonder: yes, your comment is correct, I only adding context
$endgroup$
– pasaba por aqui
Mar 19 at 21:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If we set $X=cos x$ and $Y=sin x$, the equation becomes
$$
Y=frac12X-X^2
$$
so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.
This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.
The equation becomes
$$
X^4-X^3+frac54X^2-1=0
$$
as you found out. The two real roots are approximately
$$
-0.654665139167 qquad 0.921490878816
$$
These correspond to $x=pm2.284535877184578$ and $x=pm0.39889463967156$, that correspond to what WolframAlpha finds.
answered Mar 19 at 21:47
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
If we put $t=x/2$ then we get $$2tan tover 1-tan ^2t +2cos ^2 t -1=1over 2$$
Let $y= tan t$. Since $cos ^2t = 1over 1+y^2$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
add a comment |
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%2f3154610%2fsolve-tanxcosx-1-2%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
$begingroup$
If we set $X=cos x$ and $Y=sin x$, the equation becomes
$$
Y=frac12X-X^2
$$
so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.
This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.
The equation becomes
$$
X^4-X^3+frac54X^2-1=0
$$
as you found out. The two real roots are approximately
$$
-0.654665139167 qquad 0.921490878816
$$
These correspond to $x=pm2.284535877184578$ and $x=pm0.39889463967156$, that correspond to what WolframAlpha finds.
answered Mar 19 at 21:47
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
If we set $X=cos x$ and $Y=sin x$, the equation becomes
$$
Y=frac12X-X^2
$$
so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.
This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.
The equation becomes
$$
X^4-X^3+frac54X^2-1=0
$$
as you found out. The two real roots are approximately
$$
-0.654665139167 qquad 0.921490878816
$$
These correspond to $x=pm2.284535877184578$ and $x=pm0.39889463967156$, that correspond to what WolframAlpha finds.
answered Mar 19 at 21:47
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
If we set $X=cos x$ and $Y=sin x$, the equation becomes
$$
Y=frac12X-X^2
$$
so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.
This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.
The equation becomes
$$
X^4-X^3+frac54X^2-1=0
$$
as you found out. The two real roots are approximately
$$
-0.654665139167 qquad 0.921490878816
$$
These correspond to $x=pm2.284535877184578$ and $x=pm0.39889463967156$, that correspond to what WolframAlpha finds.
answered Mar 19 at 21:47
egregegreg
185k1486206
$endgroup$
If we set $X=cos x$ and $Y=sin x$, the equation becomes
$$
Y=frac12X-X^2
$$
so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.
This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.
The equation becomes
$$
X^4-X^3+frac54X^2-1=0
$$
as you found out. The two real roots are approximately
$$
-0.654665139167 qquad 0.921490878816
$$
These correspond to $x=pm2.284535877184578$ and $x=pm0.39889463967156$, that correspond to what WolframAlpha finds.
answered Mar 19 at 21:47
egregegreg
185k1486206
answered Mar 19 at 21:47
egregegreg
185k1486206
answered Mar 19 at 21:47
egregegreg
185k1486206
answered Mar 19 at 21:47
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
$begingroup$
If we put $t=x/2$ then we get $$2tan tover 1-tan ^2t +2cos ^2 t -1=1over 2$$
Let $y= tan t$. Since $cos ^2t = 1over 1+y^2$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
add a comment |
$begingroup$
If we put $t=x/2$ then we get $$2tan tover 1-tan ^2t +2cos ^2 t -1=1over 2$$
Let $y= tan t$. Since $cos ^2t = 1over 1+y^2$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
add a comment |
$begingroup$
If we put $t=x/2$ then we get $$2tan tover 1-tan ^2t +2cos ^2 t -1=1over 2$$
Let $y= tan t$. Since $cos ^2t = 1over 1+y^2$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
If we put $t=x/2$ then we get $$2tan tover 1-tan ^2t +2cos ^2 t -1=1over 2$$
Let $y= tan t$. Since $cos ^2t = 1over 1+y^2$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
edited Mar 19 at 21:21
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
answered Mar 19 at 21:06
Maria MazurMaria Mazur
49.3k1360123
49.3k1360123
add a comment |
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%2f3154610%2fsolve-tanxcosx-1-2%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
$begingroup$
Wolfram alpha confirms that there are no 'simple' solutions to the equation (click on exact form): wolframalpha.com/input/?i=tan(x)%2Bcos(x)%3D1%2F2
$endgroup$
– Dr. Mathva
Mar 19 at 21:08
$begingroup$
Of course it is possible (cf. Ferrari's formula for the quartic, for example).
$endgroup$
– Allawonder
Mar 19 at 21:19
$begingroup$
@Allawonder: this question is in a scholar book for 16 years old, I doubt the was thiking on quartics
$endgroup$
– pasaba por aqui
Mar 19 at 21:23
$begingroup$
@pasabaporaqui I was answering your question. You had said, Is it possible...? I answered that it was.
$endgroup$
– Allawonder
Mar 19 at 21:26
$begingroup$
@Allawonder: yes, your comment is correct, I only adding context
$endgroup$
– pasaba por aqui
Mar 19 at 21:29