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.
Best Answer
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.