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
 A: Your error is that in general $\cos A < \cos B$ does not imply  that $A > B$. (If you don't see this right away, try $A = \pi$ and $B = \dfrac{3\pi}{2}$).
Instead, try the following approach: 
$ \cos x \cos 2x - \sin x \sin 2x < -\sin 6x $
$\cos 3x < -\sin 6x$
$\cos 3x + \sin 6x < 0$
$\cos 3x + 2\sin 3x\cos 3x < 0$
$\cos 3x(1+2\sin 3x) < 0$
So we need exactly one of $\cos 3x$ and $1+2\sin 3x$ to be negative. 
We know that $\cos 3x < 0$ when $\dfrac{\pi}{2}+2\pi k < 3x < \dfrac{3\pi}{2}+2\pi k$ for some integer $k$. Also, $1+2\sin 3x < 0$ when $\sin 3x < -\dfrac{1}{2}$, i.e. $\dfrac{7\pi}{6}+2\pi k < 3x < \dfrac{11\pi}{6}+2\pi k$ for some integer $k$. 
Can you finish the problem from here?
A: $$\cos { \left( 3x \right) < } -\sin { \left( 6x \right)  } \\ \cos { \left( 3x \right) < } -2\sin { \left( 3x \right) \cos { \left( 3x \right)  }  } \\ \cos { \left( 3x \right)  } \left( 1+2\sin { \left( 3x \right)  }  \right) <0$$
$$1.\cos { \left( 3x \right)  } >0\quad and\quad 1+2\sin { \left( 3x \right)  } <0\\ 2.\cos { \left( 3x \right)  } <0\quad and\quad 1+2\sin { \left( 3x \right)  } >0\\ $$
A: $$ \sin (x) \sin (2x) - \cos (x) \cos (2x) > \sin (6x) \Longleftrightarrow$$
$$ -\cos(3x) > \sin (6x) $$
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 $n \in \mathbb{Z}$
