Solve $\sin 2x=-\cos x$ I'm working on solving the problem  $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow  \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\pi }{2}-x +2\pi k \Leftrightarrow 3x=\frac{\pi }{2} +2\pi k\Leftrightarrow x_1=\frac{\pi }{6}+\frac{2\pi}{3} k\ $
$2x =\pi -(-\frac{\pi }{2}-x) +2\pi k \Leftrightarrow x_2=\frac{3\pi}{2}+2\pi k$
None of these answers are correct.
 The correct answer is $x=\frac{\pi}{2}+k\pi$ and $x=-\frac{\pi}{2}\pm\frac{\pi}{3}+2 k\pi$
I don't know what I'm doing wrong in my attempt.

Begin by using the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$ to make the equation $2\sin(x)\cos(x) = -\cos(x)$, or $2\sin(x)\cos(x) + \cos(x) = 0$. Now we can factor to obtain $\cos(x)(2\sin(x) + 1) = 0$. This means that either $\cos(x) = 0$ or $\sin(x) = -\frac{1}{2}$. Thus, the solutions are $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6},$ and so on.
