<s>[INST] How to solve $\sin\theta +\sin3\theta =0$ 
Solve the equation by first using a Sum-to-Product Formula.
$$\sin\theta +\sin3\theta =0$$

Steps I took:
$$\begin{align}0&=2\sin\frac { \theta +3\theta  }{ 2 } \cos\frac { \theta -3\theta  }{ 2 }\\
0&=2\sin\frac { 4\theta  }{ 2 } \cos\frac { -2\theta  }{ 2 } \\
0&=2\sin2\theta \cos(-\theta)\\
0&=2\sin2\theta \cos\theta\\
0&=2(2\sin\theta \cos\theta )\cos\theta\\
0&=4\sin\theta \cos^2\theta \\
0&=4\sin\theta \frac { 1+\cos2\theta  }{ 2 } \\
0&=4\sin\theta \frac { 1+1-2\sin^2\theta  }{ 2 } \\
0&=4\sin\theta \frac { 2-2\sin^{ 2 }\theta  }{ 2 } \\
0&=\frac { 8\sin\theta -8\sin^{ 2 }\theta  }{ 2 }\end{align}$$
 [/INST] As $\sin(-\theta)=-\sin\theta,$
$$\sin3\theta=-\sin\theta=\sin(-\theta)$$
$$3\theta=m\pi+(-1)^m(-\theta)$$ where $m$ is any integer
Check if $m$ is even $(=2r)$(say) and $m$ is odd $(=2r+1)$(say) one by one
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