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I am studying for a final and came across a review question that I have no idea how to do. The question is "Solve the equation $\sin(2x) + \sin(x) = 0$ on the interval $[0, 2\pi)$.

I can graph it and get the solutions $0, 2\pi/3, \pi$, and $4\pi/3$ where it crosses zero between $0$ and $2\pi$. My question is how do you do it algebraically?

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    $\begingroup$ Use $\sin 2x = 2\sin x \cos x$? $\endgroup$ – peterwhy Dec 17 '14 at 14:10
  • $\begingroup$ @peterwhy 7 upvotes in 1 minute; nice! $\endgroup$ – Ahaan S. Rungta Dec 17 '14 at 14:12
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    $\begingroup$ @peterwhy +1 for '?' $\endgroup$ – Aditya Hase Dec 17 '14 at 14:12
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OK I am going to write something else. Using sum to product formula,

$$\begin{align*} \sin 2x + \sin x &= 0 \\ 2\sin\frac{3x}2\cos\frac{x}{2} &= 0\\ \sin\frac{3x}2 &= 0&\text{or}& &\cos\frac{x}{2} &= 0\\ \end{align*}$$ Since $x\in[0,2\pi)$, $\frac{3x}2\in[0,3\pi)$ and $\frac x2\in[0,\pi)$. So the first case gives $$\frac{3x}2 = 0 \text{ or }\pi \text{ or }2\pi$$ And the second case gives

$$\frac{x}2 = \frac\pi2$$

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Rewriting the equation using the double-angle identity for $\sin$ gives $$2 \sin x \cos x + \sin x = 0,$$ and factoring gives $$(2 \cos x + 1) \sin x = 0.$$ This holds iff $$2 \cos x + 1 = 0 \qquad \text{or} \qquad \sin x = 0.$$

The solutions to the first equation are $\frac{2 \pi}{3}, \frac{4 \pi}{3}$, and the solutions to the second are $0, \pi$.

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First realize that $ \sin 2x = 2 \sin x \cos x $. Then, we have $$ 2 \sin x \cos x + \sin x = 0 \iff \left( 2 \cos x + 1 \right) \cdot \sin x = 0, $$ which is ture iff $ \cos x = - \frac {1}{2} $ or $ \sin x = 0 $. Can you solve each equation and find the family of solutions from here?

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Alternatively, one could rewrite the equation in complex form: $$\frac{1}{2i}(e^{2 i x} - e^{-2 i x}) + \frac{1}{2i}(e^{i x} - e^{-i x}) = 0.$$ Multiplying through by the (never-zero) quantity $2 i e^{2 i x}$ gives $$e^{4ix} + e^{3ix} - e^{ix} - 1 = 0. \qquad (\ast)$$ The left-hand side is just the polynomial $$z^4 + z^3 - z - 1 = (z + 1)(z^3 - 1)$$ evaluated at $e^{ix}$, so the solutions are the $x$ for which $e^{ix} = -1$, namely $x = \pi$, and those for which $e^{ix}$ is a third root of unity, namely, $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$.

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$$\sin2x=-\sin x=\sin(-x)$$

$$\implies2x=n\pi+(-1)^n(-x)$$ where $n$ is any integer

If $n$ is even, $=2m$(say) $$2x=2m\pi-x\iff x=\frac{2m\pi}3$$

If $n$ is odd, $=2m+1$(say) $$2x=(2m+1)\pi+x\iff x=(2m+1)\pi$$

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$$\sin{(2x)}+\sin{(x)}=2\sin{(x)}\cos{(x)}+\sin{(x)}=\sin{(x)}(2\cos{(x)}+1)=0\rightarrow$$

$$\sin{(x)}=0\rightarrow x=0\space and\space x=\pi$$

or

$$\cos{(x)}=\frac{-1}{2}\rightarrow x=\frac{2\pi}{3}\space and\space x=\frac{4\pi}{3}$$

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