Given:
(sinx + 1)(2sin²x - 3sinx - 2) = 0
sinx + 1 = 0
sinx = -1
Take the arcsin of both-side
x = sin⁻' (-1)
x = -90= -90 + 360 = 270 = 3π / 2
x =3π / 2
Similarly,
2sin²x - 3sinx - 2 = 0
let y = sinx
2y² - 3y - 2 = 0
Factorising the above;
2y² - 4y + y - 2 = 0
2y(y-2) + 1(y-2) = 0
(2y+1)(y-2) = 0
Put sin x back in for y so we can see what it's really supposed to be.
(2sinx + 1 )(sinx -2) = 0
Either one of the terms 2sin x + 1 or sin x - 2 can be 0 for this equation to be true. Let's solve each of them separately.
2sinx + 1 = 0
2sinx = -1
sinx = -1/2
Take the arcsin of both-side.
x= sin⁻' (-1/2)
x = - 30 = -30+360 = 330 ,
x = 7π /6 or 11π/6
Therefore, the correct option is B. x= 3π / 2 , 7π /6 , 11π/6