Given that $\sin\theta = 4\sin(\theta - 60°)$, show that $2\sqrt{3}\cos(\theta) = \sin(\theta) $ The answer I get when I try to simplify the first equation gives me $2\sin\theta - 2\sqrt{3}\cos\theta,$ which clearly is not the same as $2\sqrt{3}\cos\theta$.
Or am I missing something obvious? 
I've attached a screenshot of the question directly from the book.
Thanks in advance.

 A: You have $\sin\theta=4\sin(\theta-60^o)=2\sin\theta-2\sqrt3\cos\theta$.
Now subtract $\sin\theta$ from both sides:
$0=\sin\theta-2\sqrt3\cos\theta,$ 
or $2\sqrt3\cos\theta=\sin\theta$.
A: Given: $\sin \theta = 4\sin(\theta - 60^{\circ})$
Expand right hand side using angle sum (difference) identity:
$\sin \theta = 4(\sin\theta\cos 60^{\circ} - \cos\theta\sin 60^{\circ}) $
Note $\cos 60^{\circ} = \frac 12, \sin 60^{\circ} = \frac {\sqrt 3}2$
Simplifying the above:
$\sin \theta = 2\sin\theta - 2\sqrt 3{\cos\theta} $
Rearrange, simplify:
$2\sqrt 3{\cos\theta} = \sin\theta$ as required.
To finish up the solution, you can divide both sides by $\cos\theta$ after verifying that no solution of $\cos\theta = 0$ solves the equation (it doesn't, so the division is permitted).
Then solve $\tan\theta = 2\sqrt 3$ using a calculator. Within the stipulated range, the value you get by taking the inverse sine on your calculator will be the only valid solution (tangent is non-negative only in the first and third quadrants, only the first quadrant applies in this range).
