350 Followers problem- Trigo bash-1

I'm replacing $\theta$ with $x$ due to keyboard difficulties xD.

$2\left(\cos x+\cos2x\right)+\sin2x\left(1+2\cos x\right)=2\sin x$

$2\left(\cos x+\cos2x\right)+2\sin x\left(2\cos^2x+\cos x-1\right)=0$

$2\left(2\cos^2x+\cos x-1\right)+2\sin x\left(2\cos^2x+\cos x-1\right)=0$

$2\left(1+\sin x\right)\left(2\cos x-1\right)\left(\cos x+1\right)=0$

From the left term we get $x = -\frac{\pi}{2}$

From the middle term, we get $x = \frac{\pi}{3},-\frac{\pi}{3}$

From the right term, we get $x = -\pi,\pi$

Ignoring $x = -\frac{\pi}{2}$ , we have $4$ solutions.

Tedious but satisfying. Here we go!

$2(\cos\theta+\cos2\theta)+\sin2\theta(1+2\cos\theta)=2\sin\theta$

(Apply double-angle identities.)

$2(\cos\theta+\cos^{2}\theta-\sin^{2}\theta)+2\sin\theta\cos\theta(1+2\cos\theta)=2\sin\theta$

(Divide by 2,distribute, subtract $\sin\theta$ from both sides.)

$\cos\theta+2\cos^{2}\theta-1+\sin\theta\cos\theta+2\sin\theta\cos^{2}\theta-\sin\theta=0$

(Group terms in a convenient way.)

$(2\cos^{2}\theta+\cos\theta-1)+\sin\theta(2\cos^{2}\theta+\cos\theta-1)=0$

(Substitute with $x$ .)

$x+x\sin\theta=0$

$x(1+\sin\theta)=0$

(We see that $-\frac{\pi}{2}$ is a solution).

Now replace $\cos\theta$ with $x$ :

$2x^{2}+x-1=0$

(After solving we get $x=-1,\frac{1}{2}$ )

$\cos\theta=-1,\frac{1}{2}$

$\theta= \frac{-\pi}{2}, -\pi, \pi, \frac{-\pi}{3}, \frac{\pi}{3}$

So, the answer is $\boxed{5}$ .