1. [tex]\tan^2x+\tan x=\tan x(\tan x+1)=0\implies \begin{cases}\tan x=0\\\tan x+1=0\end{cases}[/tex]
Since [tex]\tan x=\dfrac{\sin x}{\cos x}[/tex], you will have [tex]\tan x=0[/tex] whenever [tex]\sin x=0[/tex]. This happens only when [tex]x=0,\pi,2\pi[/tex]. Meanwhile, [tex]\tan x+1=0\implies \tan x=-1[/tex], which happens when [tex]x=\dfrac{3\pi}4,\dfrac{7\pi}4[/tex].
2. [tex]\sin2a-\cos a=2\sin a\cos a-\cos a=\cos a(2\sin a-1)=0\implies\begin{cases}\cos a=0\\2\sin a-1=0\end{cases}[/tex]
You have [tex]\cos a=0[/tex] when [tex]a=\dfrac\pi2,\dfrac{3\pi}2[/tex], and [tex]2\sin a-1=0\implies \sin a=\dfrac12[/tex]. This happens when [tex]a=\dfrac\pi6,\dfrac{5\pi}6[/tex].
3. [tex]4\cos^2x-3=(2\cos x+\sqrt3)(2\cos x-\sqrt3)=0\implies\cos x=\pm\dfrac{\sqrt3}2[/tex]
This happens when [tex]x=\dfrac\pi6,\dfrac{5\pi}6,\dfrac{7\pi}6,\dfrac{11\pi}6[/tex].
4. [tex]\csc^2x-\csc x-2=(\csc x-2)(\csc x+1)=0\implies\begin{cases}\csc x-2=0\\\csc x+1=0\end{cases}[/tex]
You have [tex]\csc x-2=0\implies \csc x=2\implies \sin x=\dfrac12[/tex], which you know from (2) that the solutions are [tex]x=\dfrac\pi6,\dfrac{5\pi}6[/tex]. Meanwhile, [tex]\csc x+1=0\implies \csc x=-1\implies \sin x=-1[/tex], and this happens only when [tex]x=\dfrac{3\pi}2[/tex].
