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$$\sin(2x)=0.$$
 A: 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 $t=\tan x$
A: You can proceed either way, but to assure you that you can proceed as noted secondly, we know that 
$$ab = 0 \implies a = 0 \;\text{ or }\; b = 0\;\;\text{Not necessarily both equal to zero!}$$
So
$$(3\tan(x)+4\cot(x))\cdot \sin(2x) \implies (3\tan(x)+4\cot(x))=0\;\text{ or } \sin(2x)=0$$
A: Either way you should get the same answer because Mathematically the equations are still the same. Having said that, the way you've written down as one by one makes it a lot easier because identifying the zeros of a smaller expression is generally easier. 
A: Separating out the functions in trigonometric products is not always valid.  Note that $3\tan x + 4 \cot x$ is not continuous, which is a sign that you need to take care, because the product is not necessarily always defined.
Note also that knowing $\sin 2x=2\sin x\cos x$ tells you that the discontinuities will cancel, and that there is an underlying continuous function.
In this case the "problem points" are not zeros of the continuous function, so the answer is the same whether the function is taken as defined or undefined at those points.
Note also that using the formula for $\sin 2x$ the product comes out as $6\sin^2x+8\cos^2x=6+2\cos^2x$, and since $0\leq \cos^2 x \leq 1$ the function is clearly strictly positive $(\geq 6)$ wherever it is defined.
