Find all real numbers on the interval 0 2 pi that satisfy th

Find all real numbers on the interval [0, 2 pi) that satisfy the equation. Use radian measure 2 sin^2 x - 3 sin x = -1 The solution set is {}. (simplify your answer. Type an exact answer, using x as needed. Use a comma to separate

Solution

2sin^2 x - 3 sinx = -1

adding 1 on both sides

2sin^2 x - 3sin x + 1 = 0

let sin x = y

2y^2 - 3y + 1 = 0

(2y -1 ) ( y -1 ) = 0

y = 1/2 , 1

sin x = 1/2 , x = pi/6 , 5pi/6

sin x = 1 , x = pi/2

hence ,

solutions are : pi/6 , 5pi/6 , pi/2