Question

How do you solve `2sin^2x=cosx+1` in the interval `0<=x<=2pi`?

Answer 1

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

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

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

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

`0 = 2cosx(cosx + 1) - 1(cosx + 1)`

`0 = (2cosx - 1)(cosx + 1)`

`cosx = 1/2 and cosx = -1`

`x = pi/3, (5pi)/3 and pi`

Hopefully this helps!