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?
 A: Why square the equation? $\cos \theta = \frac{\sqrt 3}{2}$ is easier to solve. Also, include the solutions of $\sin \theta = 0$ before canceling $\sin \theta$.
A: 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 $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$ 
A: Dear PUK, you missed $\theta=0$ because you divided the equation by $\tan\theta$ in the first step which is only possible if $\tan\theta$ is nonzero and finite. That's why $\theta=0$ and $\theta=\pi$ and $\theta=-\pi$ for which $\tan\theta$ vanish must be separately checked and indeed, you will find out that the equation is satisfied because $0=0$.
On the other hand, you added the wrong solutions $\pm 5\pi/6$ because their squared cosine is $3/4$, but the cosine itself has a wrong sign, so your squaring created problems. The squaring was unnecessary, as pointed out by lhf as I was writing this sentence, but if you still want to square it, you have to check all the solutions that they work and you will find out that the $\pm 5\pi/6$ solutions don't.
A: 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.
A: $$\sqrt{3}\tan\theta=2\sin\theta$$
$$\sqrt{3}\frac{\sin\theta}{\cos\theta}=2\sin\theta$$
$$\sqrt{3}\sin\theta-2\sin\theta\cos\theta=0$$
$$\sin\theta(\sqrt3-2\cos\theta)=0$$
$$\sin\theta=0, \ \ \cos\theta=\frac{\sqrt3}{2}$$
$$\theta=k\pi , \ \ \ \theta=2k\pi\pm \frac{\pi}{6}$$
For given interval, $\theta\in[-\pi, \pi]$, we get
$$\color{blue}{\theta=-\pi, -\frac{\pi}{6}, 0, \frac{\pi}{6}, \pi}$$
