Solve $\sqrt{3}\tan\theta=2\sin\theta$ I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.
$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$
$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$
I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$. 
Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.
Where am I going wrong please?

When you square both sides of a equation you get a new equation that has all the solutions of the first but may admit other solutions. That was in the step $\sqrt{3}=2\cos\theta \Rightarrow 3=4\cos^2 \theta$.
Added: Also you can only divide your initial equation by $\sin \theta\neq 0$. But for $\theta=0$ or $\theta=\pm\pi$, $\sin\theta=\tan\theta =0$, which are solutions of your initial equation.
Added 2: In general the equation $A^n=B^n$ has all the solutions of the equation $A=B$ but may have additional ones. This is a consequence of the algebraic identity $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$ 
