2sin^2x-sinx-1=0, solve for x

2 sin² x - sin x - 1 = 0

u = sin x (substitution)

2 u² - u - 1=0

2 u² - 2 u + u - 1 = 0

2 u (u - 1) + (u - 1) = 0

(u - 1) + (2 u + 1) = 0

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

sin x - 1 = 0 or: 2 sin x + 1 = 0

sin x = 1 sin x = - 1/2

x 1 = π/2 + 2 kπ x 2 = 7π/6 + 2 kπ, x 3 = - π/6 + 2 kπ, k ∈ Z

k ∈ Z

2sin² (x) - sin (x) - 1 = 0

2sin² (x) - 2sin (x) + sin (x) - 1 = 0

2sin (x) [sin (x) ] - 2sin (x) [1] + 1[sin (x) ] - 1[1] = 0

2sin (x) [sin (x) - 1] + 1[sin (x) - 1] = 0

[2sin (x) + 1][sin (x) - 1] = 0

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

- 1 - 1 + 1 + 1

2sin (x) = - 1 sin (x) = 1

2 2 sin⁻¹[sin (x) ] = sin⁻¹ (1)

sin (x) = - 0.5 x = 90, 450

sin⁻¹[sin (x) ] = sin⁻¹ (-0.5)

x = - 30, 210