Solve the equation by first using a Sum-to-Product Formula.
$$\sin\theta +\sin3\theta =0$$
Steps I took:
$$\begin{align}0&=2\sin\frac { \theta +3\theta }{ 2 } \cos\frac { \theta -3\theta }{ 2 }\\
0&=2\sin\frac { 4\theta }{ 2 } \cos\frac { -2\theta }{ 2 } \\
0&=2\sin2\theta \cos(-\theta)\\
0&=2\sin2\theta \cos\theta\\
0&=2(2\sin\theta \cos\theta )\cos\theta\\
0&=4\sin\theta \cos^2\theta \\
0&=4\sin\theta \frac { 1+\cos2\theta }{ 2 } \\
0&=4\sin\theta \frac { 1+1-2\sin^2\theta }{ 2 } \\
0&=4\sin\theta \frac { 2-2\sin^{ 2 }\theta }{ 2 } \\
0&=\frac { 8\sin\theta -8\sin^{ 2 }\theta }{ 2 }\end{align}$$
Best Answer
As far as I see, you didn't make any mistake in your calculation. You did, however, overcomplicate things.
For one, if $2\sin(\alpha)\cos(\beta) = 0$, you can get rid of the $2$ and just write $\sin(\alpha)\cos(\beta) = 0$.
The main mistake you made, however, was when you got to $2\sin(2\theta)\cos\theta$. From there on, you should ask yourself this:
An edit answering your question if you were completely wrong:
No, you were not wrong. The only place you were wrong is when you rewrote
$$4\sin(\theta)\frac{2-2\sin^2\theta}{2}$$
Into $$\frac{8\sin\theta - 8\sin^2\theta}{2}.$$
There is a small mistake here. Other than that, you are correct, but truly horribly inefficient.
For example, in row $6$, you wrote $4\sin\theta \cos^2\theta = 0$, then you took another $4$ rows before you got to $4\sin(\theta)\frac{2-2\sin^2\theta}{2} = 0$, when, in fact, you could just replace $\cos^2\theta$ with $1-\sin^2\theta$ and get the same result.