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.
$\endgroup$ 13 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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 $\pi$
$\endgroup$ $\begingroup$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 $\theta\in\bigg\{-\pi,-\dfrac{2\pi}{3},0,\dfrac{2\pi}{3},\pi\bigg\}$ .
$\endgroup$
