Question

Find all solutions of the equations in the interval [0,2pi].

Answer 1

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 $x=\dfrac{3\pi}2$.