Find all solutions of the equation in the interval [0, 2π). sin(x/2)+cosx-1=0

Hi! To solve this equation we need to express cos(x) in terms of sin(x/2). It is a well-known formula: cos(x) = cos^2(x/2) - sin^2(x/2) = 1 - 2sin^2(x/2).
Now denote sin(x/2) as u and obtain an equation for u: u + 1 - 2u^2 - 1 = 0, or u - 2u^2 = 0, or u(1 - 2u) = 0.
This gives u = 0 or u = 1/2. Now recall that u = sin(x/2) and solve for x the equations sin(x/2) = 0 and sin(x/2) = 1/2.

There is the only solution for sin(x/2) = 0 on [0, 2 pi) , x_1 = 0 (x/2 = k pi , x = 2k pi ).
There are two solutions for sin(x/2) = 1/2 on [0, 2 pi) , x_2 = pi/3 and x_3 = (5 pi)/3.
(x/2 = pi/6 + 2k pi or x/2 = -pi/6 + 2k pi ).

So the answers are x_1 = 0 , x_2 = pi/3 and x_3 = (5 pi)/3.