Solve the equation sin(x/2) + cos(x) - 1 = 0.

Hello!
I suppose that "sinx/2+cosx-1=0" means "sin(x/2)+cos(x)-1=0" (it can also mean "(sinx)/2+cos(x)-1=0" and "sin(x/2)+cos(x-1)=0").
To solve this equation, recall the double angle formula cos(2a) = 1 - 2sin^2(a), and apply it to cos(x): cos(x) = 1 - 2sin^2(x/2). This way our equation becomes
sin(x/2) + (1 - 2sin^2(x/2)) - 1 = 0, or sin(x/2) - 2sin^2(x/2) = 0, or sin(x/2)(1 - 2sin(x/2)) = 0.
The product is zero means at least one of factors is zero, i.e. sin(x/2) = 0 or sin(x/2) = 1/2. These equations are well-known and their solutions are
x/2 = k pi, or x = 2k pi, x/2 = pi/6 + 2k pi, or x = pi/3 + 4k pi, x/2 = (5pi)/6 + 2k pi, or x = (5pi)/3 + 4k pi,
where k is any integer.
At [0, 4pi], which is a period of sin(x/2)+cos(x)-1, the solutions are 0, pi/3, (5pi)/3, 2pi and 4pi.