Mixing
Trigonometric inequality $ 3cos ^2x sin x -sin^2x <{1over 2}$
$begingroup$
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
And 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 Jan 3 at 14:56
greedoidgreedoid
38.8k114797
$endgroup$
|
show 3 more comments
$begingroup$
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
And 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 Jan 3 at 14:56
greedoidgreedoid
38.8k114797
$endgroup$
$begingroup$
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
$endgroup$
– Klangen
Jan 3 at 15:11
$begingroup$
$cos^nx = (cos x)^n$
$endgroup$
– greedoid
Jan 3 at 15:11
$begingroup$
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
$endgroup$
– Klangen
Jan 3 at 15:14
$begingroup$
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
$endgroup$
– Umberto P.
Jan 3 at 15:16
$begingroup$
Have you used Wolfram Alpha to find the closed form of the roots?
$endgroup$
– Szeto
Jan 3 at 15:22
|
show 3 more comments
$begingroup$
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
And 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 Jan 3 at 14:56
greedoidgreedoid
38.8k114797
$endgroup$
I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3cos ^2x sin x -sin^2x <{1over 2}$$
And 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 Jan 3 at 14:56
greedoidgreedoid
38.8k114797
asked Jan 3 at 14:56
greedoidgreedoid
38.8k114797
edited Jan 8 at 18:40
greedoid
asked Jan 3 at 14:56
greedoidgreedoid
38.8k114797
asked Jan 3 at 14:56
greedoidgreedoid
38.8k114797
asked Jan 3 at 14:56
greedoidgreedoid
38.8k114797
38.8k114797
$begingroup$
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
$endgroup$
– Klangen
Jan 3 at 15:11
$begingroup$
$cos^nx = (cos x)^n$
$endgroup$
– greedoid
Jan 3 at 15:11
$begingroup$
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
$endgroup$
– Klangen
Jan 3 at 15:14
$begingroup$
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
$endgroup$
– Umberto P.
Jan 3 at 15:16
$begingroup$
Have you used Wolfram Alpha to find the closed form of the roots?
$endgroup$
– Szeto
Jan 3 at 15:22
|
show 3 more comments
$begingroup$
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
$endgroup$
– Klangen
Jan 3 at 15:11
$begingroup$
$cos^nx = (cos x)^n$
$endgroup$
– greedoid
Jan 3 at 15:11
$begingroup$
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
$endgroup$
– Klangen
Jan 3 at 15:14
$begingroup$
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
$endgroup$
– Umberto P.
Jan 3 at 15:16
$begingroup$
Have you used Wolfram Alpha to find the closed form of the roots?
$endgroup$
– Szeto
Jan 3 at 15:22
$begingroup$
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
$endgroup$
– Klangen
Jan 3 at 15:11
$begingroup$
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
$endgroup$
– Klangen
Jan 3 at 15:11
$begingroup$
$cos^nx = (cos x)^n$
$endgroup$
– greedoid
Jan 3 at 15:11
$begingroup$
$cos^nx = (cos x)^n$
$endgroup$
– greedoid
Jan 3 at 15:11
$begingroup$
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
$endgroup$
– Klangen
Jan 3 at 15:14
$begingroup$
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
$endgroup$
– Klangen
Jan 3 at 15:14
$begingroup$
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
$endgroup$
– Umberto P.
Jan 3 at 15:16
$begingroup$
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
$endgroup$
– Umberto P.
Jan 3 at 15:16
$begingroup$
Have you used Wolfram Alpha to find the closed form of the roots?
$endgroup$
– Szeto
Jan 3 at 15:22
$begingroup$
Have you used Wolfram Alpha to find the closed form of the roots?
$endgroup$
– Szeto
Jan 3 at 15:22
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 Jan 3 at 16:35
OldboyOldboy
7,3461833
$endgroup$
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
add a comment |
$begingroup$
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 Jan 3 at 16:38
siroussirous
1,6091513
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 3 at 16:35
OldboyOldboy
7,3461833
$endgroup$
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
add a comment |
$begingroup$
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 Jan 3 at 16:35
OldboyOldboy
7,3461833
$endgroup$
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
add a comment |
$begingroup$
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 Jan 3 at 16:35
OldboyOldboy
7,3461833
$endgroup$
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 Jan 3 at 16:35
OldboyOldboy
7,3461833
edited Jan 3 at 16:42
answered Jan 3 at 16:35
OldboyOldboy
7,3461833
answered Jan 3 at 16:35
OldboyOldboy
7,3461833
answered Jan 3 at 16:35
OldboyOldboy
7,3461833
7,3461833
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
add a comment |
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you!
$endgroup$
– greedoid
Jan 3 at 16:44
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
$begingroup$
@greedoid It's nice to see that we are from the same country. Keep up the good work!
$endgroup$
– Oldboy
Jan 3 at 16:45
add a comment |
$begingroup$
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 Jan 3 at 16:38
siroussirous
1,6091513
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 16:38
siroussirous
1,6091513
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 16:38
siroussirous
1,6091513
$endgroup$
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 Jan 3 at 16:38
siroussirous
1,6091513
answered Jan 3 at 16:38
siroussirous
1,6091513
answered Jan 3 at 16:38
siroussirous
1,6091513
answered Jan 3 at 16:38
siroussirous
1,6091513
1,6091513
add a comment |
add a comment |
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$begingroup$
By $cos^2$ do you mean $cos(cos(x))$ or $cos(x)timescos(x)$?
$endgroup$
– Klangen
Jan 3 at 15:11
$begingroup$
$cos^nx = (cos x)^n$
$endgroup$
– greedoid
Jan 3 at 15:11
$begingroup$
Then clearly $x=(2n+1)pi$ for $ninmathbb{Z}$ is a solution.
$endgroup$
– Klangen
Jan 3 at 15:14
$begingroup$
A quick look at the graph shows that equality holds almost at $pi/17$. Wolfram shows this is not exact however.
$endgroup$
– Umberto P.
Jan 3 at 15:16
$begingroup$
Have you used Wolfram Alpha to find the closed form of the roots?
$endgroup$
– Szeto
Jan 3 at 15:22