Use inverse functions where needed to find all solutions of

Use inverse functions where needed to find all solutions of the equation in the interval [0, 2 pi). 2 sin^2 x - 19 sin x + 9 = 0 x = _____ (smaller value) x = _____ (larger value)

Solution

2sin^2x - 19sinx +9 =0

2sin^2x -18sinx -sinx +9 =0

2sinx( sinx -9) -1(sinx-9) =0

(sinx-9)(2sinx -1) =0

sinx= 9 Neglect as sinx cannot be greater than 1

sinx = 1/2

x = sin^-1(1/2) = pi/6

x = pi - sin^-1(1/2) = pi -pi/6 = 5pi/6

Solution : in interval [0, 2pi) x = pi/6 , 5pi/6