How do you solve sin^2θ−2sinθ+1=0 [0,2pi)?

Question

sin^2θ−2sinθ+1=0 [0,2pi)

Answer 1

Use substitution.

y = sinθ

y² - 2y + 1 = 0

(y - 1)² = 0

y = 1

sinθ = 1

θ = π/2