Question

How do you solve `cos^2x = root(3)(cosx)` on `[0, 2pi)`?

Answer 1

`cos^2x = root(3)(cosx)`

`(cos^2x)^3 = (root(3)(cosx))^3`

`cos^6x = cosx`

`cos^6x - cosx= 0`

`cos^5x(cosx- 1 ) = 0`

`cos^5x = 0 and cosx = 1`

`x = pi/2, (3pi)/2, 0`

Hopefully this helps!