Question

How do you find all solutions of the equation in the interval `[0,2pi)` given `cos4x(cosx-1)=0`?

Answer 1

`x=pi/8,3*pi/8,5pi/8,7pi/8,9pi/8,11pi/8,13pi/8,15pi/8` or `x=0`

Explanation:

We have to distinguish two cases:
`cos(4x)=0`
this gives
`4x=(2k+1)*pi/2`
`x=(2k+1)*pi/8`
the interval for `k` is `1\le k\le 7` and `k` is an integer.

`cos(x)=1`

here is only `x=0` one Solutions.