How to solve the equation $cos x cos 7x = cos 5x cos 3x$
$begingroup$
We were given the equation
$$cos x cos 7x = cos 5x cos 3x.$$
How can we solve it for $x$? So far I have solved only basic trigonometric equations, so I don't know how to reduce it to basic one.
Thanks in advance for some hints on how to solve the equation.
algebra-precalculus trigonometry
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
$endgroup$
add a comment |
$begingroup$
We were given the equation
$$cos x cos 7x = cos 5x cos 3x.$$
How can we solve it for $x$? So far I have solved only basic trigonometric equations, so I don't know how to reduce it to basic one.
Thanks in advance for some hints on how to solve the equation.
algebra-precalculus trigonometry
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
$endgroup$
3
$begingroup$
You can use the identity $cos A cos B=frac{1}{2}( cos(A+B)+cos(A-B))$
$endgroup$
– Ricardo Largaespada
Dec 18 '18 at 21:51
$begingroup$
Formally mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 5:26
add a comment |
$begingroup$
We were given the equation
$$cos x cos 7x = cos 5x cos 3x.$$
How can we solve it for $x$? So far I have solved only basic trigonometric equations, so I don't know how to reduce it to basic one.
Thanks in advance for some hints on how to solve the equation.
algebra-precalculus trigonometry
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
$endgroup$
We were given the equation
$$cos x cos 7x = cos 5x cos 3x.$$
How can we solve it for $x$? So far I have solved only basic trigonometric equations, so I don't know how to reduce it to basic one.
Thanks in advance for some hints on how to solve the equation.
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
edited Dec 18 '18 at 23:32
rafa11111
1,1942417
1,1942417
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
asked Dec 18 '18 at 21:47
someone123123someone123123
457414
457414
3
$begingroup$
You can use the identity $cos A cos B=frac{1}{2}( cos(A+B)+cos(A-B))$
$endgroup$
– Ricardo Largaespada
Dec 18 '18 at 21:51
$begingroup$
Formally mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 5:26
add a comment |
3
$begingroup$
You can use the identity $cos A cos B=frac{1}{2}( cos(A+B)+cos(A-B))$
$endgroup$
– Ricardo Largaespada
Dec 18 '18 at 21:51
$begingroup$
Formally mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 5:26
3
3
$begingroup$
You can use the identity $cos A cos B=frac{1}{2}( cos(A+B)+cos(A-B))$
$endgroup$
– Ricardo Largaespada
Dec 18 '18 at 21:51
$begingroup$
You can use the identity $cos A cos B=frac{1}{2}( cos(A+B)+cos(A-B))$
$endgroup$
– Ricardo Largaespada
Dec 18 '18 at 21:51
$begingroup$
Formally mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 5:26
$begingroup$
Formally mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 5:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using the linearisation identity mentioned in a comment, it is equivalent to the equation
$cos 6x=cos 2x$,. The solutions of the latter satisfy the congruences
$$6xequivpm2xmod 2pi.$$
Can you end the computations and simplify the results?
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
$endgroup$
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%2f3045751%2fhow-to-solve-the-equation-cos-x-cos-7x-cos-5x-cos-3x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the linearisation identity mentioned in a comment, it is equivalent to the equation
$cos 6x=cos 2x$,. The solutions of the latter satisfy the congruences
$$6xequivpm2xmod 2pi.$$
Can you end the computations and simplify the results?
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Using the linearisation identity mentioned in a comment, it is equivalent to the equation
$cos 6x=cos 2x$,. The solutions of the latter satisfy the congruences
$$6xequivpm2xmod 2pi.$$
Can you end the computations and simplify the results?
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Using the linearisation identity mentioned in a comment, it is equivalent to the equation
$cos 6x=cos 2x$,. The solutions of the latter satisfy the congruences
$$6xequivpm2xmod 2pi.$$
Can you end the computations and simplify the results?
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
$endgroup$
Using the linearisation identity mentioned in a comment, it is equivalent to the equation
$cos 6x=cos 2x$,. The solutions of the latter satisfy the congruences
$$6xequivpm2xmod 2pi.$$
Can you end the computations and simplify the results?
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
answered Dec 18 '18 at 21:59
BernardBernard
123k741117
123k741117
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%2f3045751%2fhow-to-solve-the-equation-cos-x-cos-7x-cos-5x-cos-3x%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
3
$begingroup$
You can use the identity $cos A cos B=frac{1}{2}( cos(A+B)+cos(A-B))$
$endgroup$
– Ricardo Largaespada
Dec 18 '18 at 21:51
$begingroup$
Formally mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 5:26