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:
$$\begin{align}0&=2\sin\frac { \theta +3\theta  }{ 2 } \cos\frac { \theta -3\theta  }{ 2 }\\
0&=2\sin\frac { 4\theta  }{ 2 } \cos\frac { -2\theta  }{ 2 } \\
0&=2\sin2\theta \cos(-\theta)\\
0&=2\sin2\theta \cos\theta\\
0&=2(2\sin\theta \cos\theta )\cos\theta\\
0&=4\sin\theta \cos^2\theta \\
0&=4\sin\theta \frac { 1+\cos2\theta  }{ 2 } \\
0&=4\sin\theta \frac { 1+1-2\sin^2\theta  }{ 2 } \\
0&=4\sin\theta \frac { 2-2\sin^{ 2 }\theta  }{ 2 } \\
0&=\frac { 8\sin\theta -8\sin^{ 2 }\theta  }{ 2 }\end{align}$$
 A: 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:

If $a\cdot b = 0$, what can I say about $a$ and $b$?


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.
A: 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
A: $$\sin(\theta) + \sin(3\theta) = 0$$
Using sum-to-product,
$$2\sin\left(\frac{\theta+3\theta}{2}\right)\cos\left(\frac{\theta - 3\theta}{2}\right) = 0$$ 
$$2\sin(2\theta)\cos(-\theta) = 0$$
$$2\sin(\theta+\theta)\cos(-\theta)= 0$$ 
Using either the double angle or sum-difference formulas,
$$4\sin(\theta)\cos(\theta)\cos(-\theta) = 0$$
By the definition of cosine and getting rid of the 4,
$$\sin(\theta)\cos^2(\theta) = 0$$
Using co-function identities,
$$\tag{1}\sin(\theta)\sin^2(\frac{\pi}{2} - \theta) = 0 $$
We know that $\sin$ is zero at $k\pi$ for $k \in \mathbb{Z}$ so,
$$\theta = k\pi$$ in which case the left-hand $\sin$ in (1) is 0 or, $$\theta = \frac{\pi}{2} + k\pi$$ in which case the right-hand $\sin$ in (1) is zero.
