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

$$\cos { \left( 3x \right) < } -\sin { \left( 6x \right)  } \\ \cos { \left( 3x \right) < } -2\sin { \left( 3x \right) \cos { \left( 3x \right)  }  } \\ \cos { \left( 3x \right)  } \left( 1+2\sin { \left( 3x \right)  }  \right) <0$$
$$1.\cos { \left( 3x \right)  } >0\quad and\quad 1+2\sin { \left( 3x \right)  } <0\\ 2.\cos { \left( 3x \right)  } <0\quad and\quad 1+2\sin { \left( 3x \right)  } >0\\ $$
