Question

Directions: Find all solutions of the equation in the interval (0, 2pi)


sin x/2=1-cosx

Answer 1

recall that
cos (2x) = 1 - 2sin^2 x
or
cos x = 1 - 2sin^2 (x/2)

then yours is ...
sin (x/2) = 1 - (1 - 2sin^2 (x/2))

sin (x/2) = 2sin^2 (x/2)
2sin^2 (x/2) - sin (x/2) = 0
sin x/2(2sin x/2 - 1) = 0

sin (x/2) = 0 or sin(x/2) = 1/2

case 1: sin (x/2) = 0
x/2 = 0 , π, 2π
x = 0, 2π, 4π

case 2: sin (x/2) = 1/2
x/2 = π/6 or 5π/6 , ( 30° or 150°)
x = π/3 , 5π/3

so x = 0, , π/3, 5π/3, 2π for 0 ≤ x ≤ 2π