Find all 0 ≤ x < 2π such that 2cos^2 x + sin x − 1 = 0.

Answer 1

2cos^2(x) + sinx - 1 =0

2(1-sin^2(x)) + sinx -1 =0

2-2sin^2(x) + sinx -1 =0

-2sin^2(x) + sinx +1 =0

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

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

sin(x) = 1, -1/2

x = 90, 210, 330

= pi/2, 7pi/6, 11pi/6

2cos^2(90) + sin(90) -1 = 2(0)+1-1 = 0

2cos^2(210) + sin(210) -1 = 2(-sqr3/2)^2-1/2-1 = 3/2-3/2=0

2cos^2(330) + sin(330)-1 = 2(sqr3/2)^2-1/2-1 = 3/2-3/2=0

Answer 2

Use the identity cos²(x) = 1 - sin²(x) to get the equation 2 - 2sin²(x) + sin(x) - 1 = 0 or 2sin²(x) - sin(x) - 1 = 0. This is a quadratic in sin(x). Factor as (2sin(x) + 1)(sin(x) - 1) = 0 to get the two equations sin(x) = -1/2 or sin(x) = 1. In [0, 2π) this gives x = 7π/6, 11π/6, π/2.