# 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 $$k\in \mathbb{Z}$$  • By $\cos^2$ do you mean $\cos(\cos(x))$ or $\cos(x)\times\cos(x)$? – Klangen Jan 3 at 15:11
• $\cos^nx = (\cos x)^n$ – Aqua Jan 3 at 15:11
• Then clearly $x=(2n+1)\pi$ for $n\in\mathbb{Z}$ is a solution. – Klangen Jan 3 at 15:14
• A quick look at the graph shows that equality holds almost at $\pi/17$. Wolfram shows this is not exact however. – Umberto P. Jan 3 at 15:16
• Have you used Wolfram Alpha to find the closed form of the roots? – Szeto Jan 3 at 15:22

## 2 Answers

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.

• WelI, I took it from Tangenta-86/2 year 2016/17. You can see it there on page 43 and 44. Thank you! – Aqua Jan 3 at 16:44
• @greedoid It's nice to see that we are from the same country. Keep up the good work! – Oldboy Jan 3 at 16:45

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