Find zeros of this function: $$(3\tan(x)+4\cot(x))\cdot\sin(2x)$$
Do I have to multiply them and solve, or one by one, like:
$$(3\tan(x)+4\cot(x))=0$$and$$\sin(2x)=0.$$

Observe that 
if $\displaystyle(3\tan x+4\cot x)=0\implies 3\tan x+\frac4{\tan x}=0\iff3\tan^2x+4=0$ which is impossible for real $x$
If $\sin2x=0, 2x=n\pi$ where $n$ is any integer
If $n$ is even $=2m$(say) $2x=2m\pi, x=m\pi,\cot x=\cot m\pi=\frac{\cos m\pi}{\sin m\pi}=\frac{(-1)^m}0$ hence not finite
If $n$ is odd $=2m+1$(say) $2x=(2m+1)\pi, x=\frac{(2m+1)\pi}2,\tan x=\tan\frac{(2m+1)\pi}2=\frac{(-1)^m}0$ hence not finite
So, we don't have any real solution which will be more evident below
Method $1:$
On multiplication, 
$\displaystyle(3\tan x+4\cot x)\sin2x=\left(3\frac{\sin x}{\cos x}+4\frac{\cos x}{\sin x}\right)2\sin x\cos x=6\sin^2x+8\cos^2x$
Using $\cos2A=2\cos^2A-1=1-2\sin^2A$
$6\sin^2x+8\cos^2x=3(1-\cos2x)+4(1+\cos2x)=7+\cos2x$
Do you know for real $y,-1\le \cos y\le 1$
Method $2:$
Using $\displaystyle\sin2A=\frac{2\tan x}{1+\tan^2x}$ and $\displaystyle\cot x=\frac1{\tan x}$
$\displaystyle(3\tan x+4\cot x)\sin2x=\left(3t+\frac4t\right)\frac{2t}{1+t^2}=\frac{2(3t^2+4)}{1+t^2}$ where $t=\tan x$
