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.
You are almost done. I only evaluate the case $\cos3\theta$. You can easily check the remainder case.
You get $\cos3\theta = \cos^3\theta - 3\cos\theta \sin^2\theta$. We can apply the identity $\sin^2\theta = 1-\cos^2\theta$. Substitute it gives the formula you want.
Best Answer
It's $$2\cos2\theta\cos\theta-3\cos2\theta=2\sin2\theta\cos\theta-3\sin2\theta$$ or $$(2\cos\theta-3)(\cos2\theta-\sin2\theta)=0$$ or $$\tan2\theta=1.$$ Can you end it now?