Question

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

4 sin2x - 4 sin x + 1 = 0

Answer 1

Answer:

x = π/6, 5π/6

Step-by-step explanation:

4 sin² x − 4 sin x + 1 = 0

(2 sin x − 1)² = 0

sin x = ½

x = π/6, 5π/6

Answer 2

Answer:

x = π/6, 5π/6

Step-by-step explanation:

4 sin²x - 4 sin x + 1 = 0

(2 sin x − 1)² = 0

sinx = 1/2

using unit circle , when sinx =y= 1/2, there are two possible answers in (0,2π) range

x = π/6, 5π/6