Find the solution in [0,2pi)

Question

A) 2sin² x=1

Answer 1

2sin² x=1

sin² x=1/2

sin x = +- 1/Sqrt(2)

x = +- pi/4

Answer 2

First get the trig function isolated:

2sin²x = 1

sin²x = 1/2

sinx = +-√(1/2) = +-√2/2

sinx = +√2/2 would have solutions in Quadrants I and II. The reference angle is π/4, so the 2 solutions here are x = π/4 and x = π - π/4 = 3π/4

sinx = -√2/2 would have solutions in Quadrants III and IV. The 2 solutions here are x = 2π - π/4 = 7π/4 and x = π + π/4 = 5π/4.

So the solutions are: x = π/4, 3π/4, 5π/4, and 7π/4.