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

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



 A: 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.
A: 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 $x=\frac{-3\pi}{14}$.
