Question

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

1. tan^(2) x + tan x = 0
2. sin 2 a - cos a = 0
3. 4 cos^(2) x - 3 = 0
4. csc^(2) x - csc x - 2 = 0

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=(\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=\frac{3\pi}4,\frac{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=\frac\pi2,\frac{3\pi}2`, and
`2\sin a-1=0\implies \sin a=\frac12`. This happens when
`a=\frac\pi6,\frac{5\pi}6`.

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

This happens when
`x=\frac\pi6,\frac{5\pi}6,\frac{7\pi}6,\frac{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=\frac12`, which you know from (2) that the solutions are
`x=\frac\pi6,\frac{5\pi}6`. Meanwhile,
`\csc x+1=0\implies \csc x=-1\implies \sin x=-1`, and this happens only when
`x=\frac{3\pi}2`.