Find all solutions of the equation in the interval 0 2pi 2si

Find all solutions of the equation in the interval [0, 2pi). 2sin^2 x - 3 sinx + 1 = 0 Write your answer in radians in terms of pi. If there is more than one solution, separate them with commas. x =

Solution

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

factoring the equation

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

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

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

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

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