Question

Solve the equation on the interval 0<x<2pi Cos^2x-sin^2x=1-sinx What is the whole solutions set?

Using double angles I got Cos2x=1-sinx
but don't know what to do next.

Answer 1

`0, pi, 2pi, pi/6 and (5pi)/6`

Explanation:

f(x) = cos^2 x - sin^2 x - 1 + sin x = 0
Replace `cos^2 x by (1 - sin^2 x)` -->
`f(x) = 1 - sin^2 x - sin^2 x - 1 + sin x = 0`
`f(x) = -2sin^2 x + sin x = 0`
`f(x) = sin x(-2sin x + 1) = 0`
Two solutions:
`sin x = 0` --> x = 0 and `x = pi` and `x = 2pi`
`sin x = 1/2` --> `x = pi/6` and `x = (5pi)/6`
`x = 0, pi, 2pi, pi/6 and (5pi)/6`