Solving $2\sin\theta\cos\theta + \sin\theta = 0$
From your first line to your second you did not square correctly. If you are squaring the equation you missed the cross term $4\sin^2 \theta \cos \theta$. If you took the $\sin \theta$ to the other side before squaring to avoid the cross term, when you brought it back the $\sin^2 \theta$ should have a minus sign.
For $-\pi\leq\theta\leq\pi$: \begin{align*} 2\sin{\theta}\cos{\theta}+\sin{\theta}&=0 \\ \sin{\theta}\big(2\cos{\theta}+1\big)&=0 \\ \end{align*} Thus, $\sin{\theta}=0$ or $2\cos{\theta}+1=0$ .
- If $\sin{\theta}=0$, then $\theta\in\{-\pi,0,\pi\}$ .
- If $2\cos{\theta}+1=0$, then $\cos{\theta}=-1/2<0$, then $\theta\in(-\pi,-\pi/2)\cup(\pi/2,\pi)$ and thus $\theta\in\{-2\pi/3,2\pi/3\}$
The solution is $\theta\in\bigg\{-\pi,-\dfrac{2\pi}{3},0,\dfrac{2\pi}{3},\pi\bigg\}$ .
I'll start by graphing this function x-axis is $\theta / \pi $ which shows the function is zero at 5 points.
\begin{align} 2 \cdot \sin(\theta)\cos(\theta) + \sin(\theta) & = 0 \\ sin(\theta) \cdot (2\cdot\cos(\theta)+1) & = 0 \end{align}
So either $\sin(\theta) = 0$ or $2\cdot\cos(\theta)+1 = 0 \Rightarrow \cos(\theta) = -0.5$
Considering each of these cases then $\theta$ is $-\pi$, $-\dfrac{2}{3}\pi$, $0$, $\dfrac{2}{3}\pi$ or $\pi$