Solve the equation over the interval 02pi Solve the equation

Solve the equation over the interval [0,2pi)

Solve the equation over the interval [0, 2pi) 2 sin^2 x - sin x - 1 = 0

Solution

2sin²x-sinx-1=0

2sin²x -2sinx +sinx -1=0

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

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

sinx=-1/2        sinx=1

x=sin^-1(-1/2)          x=sin^-11

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

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