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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.

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    $\begingroup$ How did you square? $\endgroup$ – Math-fun Feb 1 at 15:06
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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.

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  • $\begingroup$ Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3\sin^2 - 4\sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot. $\endgroup$ – Joseph Feb 1 at 15:47
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    $\begingroup$ The arcsine is defined in the range $[-90^\circ,90^\circ]$ to make it a function. You know you need some angle that has a sine of $\frac {\sqrt 3}2$. Both $60^\circ$ and $120^\circ$ satisfy that. You introduced $60^\circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant. $\endgroup$ – Ross Millikan Feb 1 at 15:51
  • $\begingroup$ Thanks a lot for your help Ross, that was incredibly useful. $\endgroup$ – Joseph Feb 15 at 18:13
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I'll start by graphing this function x-axis is $\theta / \pi $ which shows the function is zero at 5 points.

Graph

\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 $\pi$

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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 $2\cos{\theta}+1=0$ .

  1. If $\sin{\theta}=0$, then $\theta\in\{-\pi,0,\pi\}$ .
  2. 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 $\theta\in\bigg\{-\pi,-\dfrac{2\pi}{3},0,\dfrac{2\pi}{3},\pi\bigg\}$ .

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