Question

How to do this?

Answer 1

`0,(2pi)/3, (4pi)/3`

Explanation:

Using identity:

`cos(2x)=2cos^2x-1`

`2cos^2x-1-cosx=0`

Let `cosx=m`

`2m^2-1-m=0`

Factor:

`(2m+1)(m-1)=0=>m=1 and m=-1/2`

Substitute `m=cosx`

`cosx=1`

`cosx=-1/2`

`cosx=1=>x=0`

`cosx=-1/2=>x=(2pi)/3, (4pi)/3`

Answer 2

`0; (2pi)/3; (4pi)/3; pi`

Explanation:

cos 2x - cos x = 0
Replace cos 2s by `(2cos^2 x - 1)`
`2cos^2 x - cos x - 1 = 0.`
Solve this quadratic equation for cos x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
cos x =1 and `cos x = c/a = - 1/2`
a. cos x = 1
Unit circle gives 2 solutions:
x = 0, and `x = 2pi`
b. `cos x = -1/2`
Trig table and unit circle give 2 solutions
`x = +- (2pi)/3`
The arc `- (2pi)/3` is co-terminal to arc `(4pi)/3`
Answers for `[0, 2pi]`:
`0; (2pi)/3; (4pi)/3; 2pi`