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.

1 Answer
Feb 15, 2016

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