Question

Find all solutions in the interval [0, 2π).

4 sin2x - 4 sin x + 1 = 0

Answer 1

@thomaster @dan815 @nincompoop @Preetha

Answer 2

Solve for x on 0 ≤ x ≤2π ;
4sin² x – 4sin x +1 = 0
(2sin x – 1)² = 0
(2sin x – 1) = 0
2sin x = 1
sin x = 1/2
x = sinˉ¹(1/2) = {π/6, 5π/6} radians = {30º, 150º}

Answer 3

Solve for x on 0 ≤ x ≤2π ;
4sin² x – 4sin x +1 = 0
(2sin x – 1)² = 0
(2sin x – 1) = 0
2sin x = 1
sin x = 1/2
x = sinˉ¹(1/2) = {π/6, 5π/6} radians = {30º, 150º}