Solving $2\sin\theta\cos\theta + \sin\theta = 0$ 
The question is to solve the following question in the range $-\pi \le \theta \le  \pi$
$$2\sin\theta\cos\theta + \sin\theta = 0$$

I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $\pm2/3\pi$ and the values when $\sin\theta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$\begin{align}
2\sin\theta\cos\theta + \sin\theta &= 0 \qquad\text{(square)} \tag{1} \\
4\sin^2\theta\cos^2\theta + \sin^2\theta &= 0 \tag{2}\\
4\sin^2\theta(1-\sin^2\theta) + \sin^2\theta &= 0 \tag{3} \\
4\sin^2\theta - 4\sin^4\theta + \sin^2\theta &= 0 \tag{4} \\
5\sin^2\theta  - 4\sin^4\theta &= 0 \tag{5}
\end{align}$$
and then solving by substitution/the quadratic equation I get $\sin\theta = \pm\sqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.

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 $\theta\in\bigg\{-\pi,-\dfrac{2\pi}{3},0,\dfrac{2\pi}{3},\pi\bigg\}$ .
