$$\sin(3a)=2 \sin(a)$$
What if I first did was split $\sin(3a)$ using the addition formula, thus giving me:
$\sin(2a)\cos(a)+\cos(2a)\sin(a)$ $\Rightarrow$
Then I split the double angles using the double angle formulas.
$2\sin(a)\cos^2(a)+(1-2\sin^2(a))\sin(a)=2\sin(a)$
Then I simplified that and got:
$2\sin(a)(\cos^2(a)-\sin^2(a))=\sin(a) \Rightarrow \cos(2a)=1/2$
This is only one answer, and I don't know if it is correct. Any hints?
On
This is what I got (and feel free to correct me if I'm wrong!)
Use the identity $\sin 3a = 3 \sin a - 4 \sin^3 a$ and set equal to $2 \sin a$ to get
$$3 \sin a - 4 \sin^3 a = 2 \sin a$$
Bring over the $2 \sin a$ to get
$$\sin a - 4 \sin^3 a = 0$$
Factor out $\sin a$ to get
$$\sin a (1-4 \sin^2 a) = 0$$
Now for a little trick...using $\cos 2a = 1-2 \sin^2 a$, we can rewrite $1-4 \sin^2 a$ as:
$$2 - 4 \sin^2 a = 2 \cos 2a \Rightarrow 1-4 \sin^2 a = 2 \cos 2a -1 $$ to get
$$\sin a (2 \cos 2a -1) = 0$$
We have either $\sin a = 0$ or $(2\cos 2a - 1) = 0$, but as we're interested in $\cos 2a$ we have
$$2 \cos 2a = 1$$
Now all we need to do is divide by $2$ to get
$$\bbox [white,5px, border:2px solid black] {\cos 2a = \dfrac {1}{2}}$$
Your solution is correct except for a small problem.
Note that when you cancelled $\sin (\alpha)$ from both sides you have to make sure to add the solutions of $\sin (\alpha)=0$ as well.
Also, you could have used the identity, $$2\cos ^2 (\alpha ) = 1+ \cos (2\alpha)$$ to have a shorter proof, but what you did in just fine.