<s>[INST] Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$  \cos x \cos 2x -   \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get: 
$$ 3x > \dfrac{\pi}{2}+6x+2k\pi$$
$$ -3x > \dfrac{\pi}{2}+2k\pi$$
$$ x < -\dfrac{\pi}{6}+ \dfrac{2k\pi}{3}$$
and
$$ 3x < -\dfrac{\pi}{2} - 6x + 2k\pi$$
$$ 9x < -\dfrac{\pi}{2} + 2k\pi$$
$$ x < - \dfrac{\pi}{18} + \dfrac{2k\pi}{9} $$
as you can see the solution is not correct
 [/INST] $$ \sin (x) \sin (2x) - \cos (x) \cos (2x) > \sin (6x) \Longleftrightarrow$$
$$ -\cos(3x) > \sin (6x) $$
So we find $4$ solutions:
$$x=\frac{2\pi n +\pi}{3}$$
$$\frac{2\pi n -\pi}{3}<x<\frac{12\pi n -5\pi}{18}$$
$$\frac{4\pi n -\pi}{6}<x<\frac{12\pi n -\pi}{18}$$
$$\frac{4\pi n +\pi}{6}<x<\frac{2\pi n +\pi}{3}$$
With $n \in \mathbb{Z}$
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