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 cancelled the $\sin \theta$ factors, you lost some solutions. This is similar to starting with $x^2 = x$, cancelling an $x$ and concluding that $x=1$; the obvious, $x=0$ solution has been lost. The remedy for this is to factor instead. In the example I provided, that would amount to writing $x^2 - x = 0$, then factoring to get $x(x-1)=0$ then setting each factor equal to zero, as usual.
You also gained solutions when you squared. This is similar to starting with $x=1$, then squaring to get $x^2 = 1$, so $x= \pm 1$. Notice that in your problem, squaring was superfluous since a few steps later you took a square root that "undid" that squaring anyway.
