Trigonometric inequality $ 3cos ^2x sin x -sin^2x <{1over 2}$
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
with no succes. Of course if we write $s=sin x$ then $cos^2 x = 1-s^2$ and we get $$6s^3+2s^2-6s+1>0$$
But this one has no rational roots so here stops. I suspect that Cardano wasn't in a mind of a problem proposer. There must be some trigonometric trick I don't see. I also tried with $$sin 3x = -4s^3+3s$$ but don't now what to do with this. Any idea?
Offical solution is a union of $({(12k-7)pi over 18},{(12k+1)piover 18})$ where $kin mathbb{Z}$
trigonometry
asked yesterday
greedoid
38.2k114797
|
show 3 more comments
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
with no succes. Of course if we write $s=sin x$ then $cos^2 x = 1-s^2$ and we get $$6s^3+2s^2-6s+1>0$$
But this one has no rational roots so here stops. I suspect that Cardano wasn't in a mind of a problem proposer. There must be some trigonometric trick I don't see. I also tried with $$sin 3x = -4s^3+3s$$ but don't now what to do with this. Any idea?
Offical solution is a union of $({(12k-7)pi over 18},{(12k+1)piover 18})$ where $kin mathbb{Z}$
trigonometry
asked yesterday
greedoid
38.2k114797
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
– Klangen
yesterday
$cos^nx = (cos x)^n$
– greedoid
yesterday
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
– Klangen
yesterday
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
– Umberto P.
yesterday
Have you used Wolfram Alpha to find the closed form of the roots?
– Szeto
yesterday
|
show 3 more comments
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
with no succes. Of course if we write $s=sin x$ then $cos^2 x = 1-s^2$ and we get $$6s^3+2s^2-6s+1>0$$
But this one has no rational roots so here stops. I suspect that Cardano wasn't in a mind of a problem proposer. There must be some trigonometric trick I don't see. I also tried with $$sin 3x = -4s^3+3s$$ but don't now what to do with this. Any idea?
Offical solution is a union of $({(12k-7)pi over 18},{(12k+1)piover 18})$ where $kin mathbb{Z}$
trigonometry
asked yesterday
greedoid
38.2k114797
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
with no succes. Of course if we write $s=sin x$ then $cos^2 x = 1-s^2$ and we get $$6s^3+2s^2-6s+1>0$$
But this one has no rational roots so here stops. I suspect that Cardano wasn't in a mind of a problem proposer. There must be some trigonometric trick I don't see. I also tried with $$sin 3x = -4s^3+3s$$ but don't now what to do with this. Any idea?
Offical solution is a union of $({(12k-7)pi over 18},{(12k+1)piover 18})$ where $kin mathbb{Z}$
trigonometry
trigonometry
asked yesterday
greedoid
38.2k114797
asked yesterday
greedoid
38.2k114797
edited yesterday
asked yesterday
greedoid
38.2k114797
asked yesterday
greedoid
38.2k114797
asked yesterday
greedoid
38.2k114797
38.2k114797
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
– Klangen
yesterday
$cos^nx = (cos x)^n$
– greedoid
yesterday
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
– Klangen
yesterday
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
– Umberto P.
yesterday
Have you used Wolfram Alpha to find the closed form of the roots?
– Szeto
yesterday
|
show 3 more comments
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
– Klangen
yesterday
$cos^nx = (cos x)^n$
– greedoid
yesterday
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
– Klangen
yesterday
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
– Umberto P.
yesterday
Have you used Wolfram Alpha to find the closed form of the roots?
– Szeto
yesterday
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
– Klangen
yesterday
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
– Klangen
yesterday
$cos^nx = (cos x)^n$
– greedoid
yesterday
$cos^nx = (cos x)^n$
– greedoid
yesterday
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
– Klangen
yesterday
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
– Klangen
yesterday
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
– Umberto P.
yesterday
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
– Umberto P.
yesterday
Have you used Wolfram Alpha to find the closed form of the roots?
– Szeto
yesterday
Have you used Wolfram Alpha to find the closed form of the roots?
– Szeto
yesterday
|
show 3 more comments
2 Answers
2
active
oldest
votes
The solution and the problem do not match. If you define:
$$f(x)=3cos ^2x sin x -sin^2x$$
You would expect to see:
$$f(pi/18)=1/2$$
Actually it's 0.475082. So the problem and the solution do not match. But if you edit the problem just a little bit:
$$ 3cos ^2x sin x -sin^3x <{1over 2}$$
...the solution and the problem seem to be matching (note the cube instead of square in the second term on the left).
So we have a typo here! :) And the correct version of the problem is likely easier.
answered yesterday
Oldboy
7,1191832
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
add a comment |
We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
Then we have:
$sin x=cos frac{pi}{7}=sin (frac{pi}{2}-frac {pi}{7})⇒ x=frac{5pi}{14}$
Similarly $x=frac{pi}{14}$ and $x=frac{-3pi}{14}$.
answered yesterday
sirous
1,6091513
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%2f3060627%2ftrigonometric-inequality-3-cos-2x-sin-x-sin2x-1-over-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
The solution and the problem do not match. If you define:
$$f(x)=3cos ^2x sin x -sin^2x$$
You would expect to see:
$$f(pi/18)=1/2$$
Actually it's 0.475082. So the problem and the solution do not match. But if you edit the problem just a little bit:
$$ 3cos ^2x sin x -sin^3x <{1over 2}$$
...the solution and the problem seem to be matching (note the cube instead of square in the second term on the left).
So we have a typo here! :) And the correct version of the problem is likely easier.
answered yesterday
Oldboy
7,1191832
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
add a comment |
The solution and the problem do not match. If you define:
$$f(x)=3cos ^2x sin x -sin^2x$$
You would expect to see:
$$f(pi/18)=1/2$$
Actually it's 0.475082. So the problem and the solution do not match. But if you edit the problem just a little bit:
$$ 3cos ^2x sin x -sin^3x <{1over 2}$$
...the solution and the problem seem to be matching (note the cube instead of square in the second term on the left).
So we have a typo here! :) And the correct version of the problem is likely easier.
answered yesterday
Oldboy
7,1191832
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
add a comment |
The solution and the problem do not match. If you define:
$$f(x)=3cos ^2x sin x -sin^2x$$
You would expect to see:
$$f(pi/18)=1/2$$
Actually it's 0.475082. So the problem and the solution do not match. But if you edit the problem just a little bit:
$$ 3cos ^2x sin x -sin^3x <{1over 2}$$
...the solution and the problem seem to be matching (note the cube instead of square in the second term on the left).
So we have a typo here! :) And the correct version of the problem is likely easier.
answered yesterday
Oldboy
7,1191832
The solution and the problem do not match. If you define:
$$f(x)=3cos ^2x sin x -sin^2x$$
You would expect to see:
$$f(pi/18)=1/2$$
Actually it's 0.475082. So the problem and the solution do not match. But if you edit the problem just a little bit:
$$ 3cos ^2x sin x -sin^3x <{1over 2}$$
...the solution and the problem seem to be matching (note the cube instead of square in the second term on the left).
So we have a typo here! :) And the correct version of the problem is likely easier.
answered yesterday
Oldboy
7,1191832
edited yesterday
answered yesterday
Oldboy
7,1191832
answered yesterday
Oldboy
7,1191832
answered yesterday
Oldboy
7,1191832
7,1191832
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
add a comment |
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
– greedoid
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
@greedoid It's nice to see that we are from the same country. Keep up the good work!
– Oldboy
yesterday
add a comment |
We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
Then we have:
$sin x=cos frac{pi}{7}=sin (frac{pi}{2}-frac {pi}{7})⇒ x=frac{5pi}{14}$
Similarly $x=frac{pi}{14}$ and $x=frac{-3pi}{14}$.
answered yesterday
sirous
1,6091513
add a comment |
We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
Then we have:
$sin x=cos frac{pi}{7}=sin (frac{pi}{2}-frac {pi}{7})⇒ x=frac{5pi}{14}$
Similarly $x=frac{pi}{14}$ and $x=frac{-3pi}{14}$.
answered yesterday
sirous
1,6091513
add a comment |
We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
Then we have:
$sin x=cos frac{pi}{7}=sin (frac{pi}{2}-frac {pi}{7})⇒ x=frac{5pi}{14}$
Similarly $x=frac{pi}{14}$ and $x=frac{-3pi}{14}$.
answered yesterday
sirous
1,6091513
We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=cos frac{pi}{7}, cos frac{3pi}{7}, cos frac{5pi}{7}$
Then we have:
$sin x=cos frac{pi}{7}=sin (frac{pi}{2}-frac {pi}{7})⇒ x=frac{5pi}{14}$
Similarly $x=frac{pi}{14}$ and $x=frac{-3pi}{14}$.
answered yesterday
sirous
1,6091513
answered yesterday
sirous
1,6091513
answered yesterday
sirous
1,6091513
answered yesterday
sirous
1,6091513
1,6091513
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3060627%2ftrigonometric-inequality-3-cos-2x-sin-x-sin2x-1-over-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
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
– Klangen
yesterday
$cos^nx = (cos x)^n$
– greedoid
yesterday
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
– Klangen
yesterday
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
– Umberto P.
yesterday
Have you used Wolfram Alpha to find the closed form of the roots?
– Szeto
yesterday