2sin^2x = sinx + 1; 0<or=x<or=2pi?

Here 2 sin ² x = sin x + 1

2 sin ² x - sin x - 1 = 0.

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

2 sin x + 1 = 0 AND sin x - 1 =0

so

sin x = -1/2 and sin x = 1

so

x = π/2 (in case of sin x = 1)

and

x= 7π/6, 11π/6 (or you can write it as - π/6 .) [ in case of sin x = -1/2]

2sin^2x = sinx + 1

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

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

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

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

sin(x) = - 1/2

x = - π/6. 7π/6, π/2, answer//

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

sinx = -1/2 and 1

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