Trigonometric inequality $ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$

Question

I' m trying to solve this one. Find all $x$ for which following is valid:

$$ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$$

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?

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Offical solution is a union of $({(12k-7)\pi \over 18},{(12k+1)\pi\over 18})$ where $k\in \mathbb{Z}$

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Answer 1

The solution and the problem do not match. If you define:

$$f(x)=3\cos ^2x \sin x -\sin^2x$$

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:

$$ 3\cos ^2x \sin x -\sin^3x <{1\over 2}$$

...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.

Answer 2

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: $s=\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}, \cos \frac{5\pi}{7}$

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 $s=\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}, \cos \frac{5\pi}{7}$

Then we have:

$\sin x=\cos \frac{\pi}{7}=\sin (\frac{\pi}{2}-\frac {\pi}{7})⇒ x=\frac{5\pi}{14}$

Similarly $x=\frac{\pi}{14}$ and $x=\frac{-3\pi}{14}$.