Question

Find all the solutions of the given equations in the interval [0,2pi) tan^3x=tanx and cos 3x=-cos3x

Answer 1

`\tan^3x=\tan x`

`\tan^3x-\tan x=0`

`\tan x(\tan^2x-1)=0`

`\tan x(\tan x-1)(\tan x+1)=0`

`\begin{cases}\tan x=0\\\tan x-1=0\\\tan x+1=0\end{cases}`


`\tan x=(\sin x)/(\cos x)=0\implies \sin x=0\implies x=0,\pi`


`\tan x-1=0\iff\tan x=1\implies x=\frac\pi4,\frac{5\pi}4`


`\tan x+1=0\iff\tan x=-1\implies x=\frac{3\pi}4,\frac{7\pi}4`

- - -


`\cos3x=-\cos3x`

`2\cos3x=0`

`\cos3x=0\implies 3x=\frac\pi2,3x=\frac{3\pi}2\implies x=\frac\pi6,x=\frac\pi2`

This doesn't account for all the solutions, however; there are some values of
`x` that push
`3x` outside the interval
`[0,2\pi)`, so let's take a few more:


`3x=\frac{5\pi}2\implies x=\frac{5\pi}6<2\pi`

`3x=\frac{7\pi}2\implies x=\frac{7\pi}6<2\pi`

`3x=\frac{9\pi}2\implies x=\frac{9\pi}6<2\pi`

`3x=\frac{11\pi}2\implies x=\frac{11\pi}6<2\pi`

We can stop there, since the next candidate gives


`3x=\frac{13\pi}2\implies x=\frac{13\pi}6>2\pi`