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  1. \(\tan^2x+\tan x=\tan x(\tan x+1)=0\implies \begin{cases}\tan x=0\\\tan x+1=0\end{cases}\)

    Since \(\tan x=\dfrac{\sin x}{\cos x}\), you will have \(\tan x=0\) whenever \(\sin x=0\). This happens only when \(x=0,\pi,2\pi\). Meanwhile, \(\tan x+1=0\implies \tan x=-1\), which happens when \(x=\dfrac{3\pi}4,\dfrac{7\pi}4\).

    2. \(\sin2a-\cos a=2\sin a\cos a-\cos a=\cos a(2\sin a-1)=0\implies\begin{cases}\cos a=0\\2\sin a-1=0\end{cases}\)

    You have \(\cos a=0\) when \(a=\dfrac\pi2,\dfrac{3\pi}2\), and \(2\sin a-1=0\implies \sin a=\dfrac12\). This happens when \(a=\dfrac\pi6,\dfrac{5\pi}6\).

    3. \(4\cos^2x-3=(2\cos x+\sqrt3)(2\cos x-\sqrt3)=0\implies\cos x=\pm\dfrac{\sqrt3}2\)

    This happens when \(x=\dfrac\pi6,\dfrac{5\pi}6,\dfrac{7\pi}6,\dfrac{11\pi}6\).

    4. \(\csc^2x-\csc x-2=(\csc x-2)(\csc x+1)=0\implies\begin{cases}\csc x-2=0\\\csc x+1=0\end{cases}\)

    You have \(\csc x-2=0\implies \csc x=2\implies \sin x=\dfrac12\), which you know from (2) that the solutions are \(x=\dfrac\pi6,\dfrac{5\pi}6\). Meanwhile, \(\csc x+1=0\implies \csc x=-1\implies \sin x=-1\), and this happens only when \(x=\dfrac{3\pi}2\).

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