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