Find the solutions of the equation....

Question

Find the solutions of the equation that are in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)

 

2 sin² u = −1 − 3 sin u

Answer 1

2sin²u = -1 - 3sinu

2sin²u + 3sinu + 1 = 0   factor by FOIL

(2sinu + 1)(sinu + 1) = 0

 

2sinu = -1

sinu = -1/2  third and fourth quadrants

 

u = 7π/6

u = 11π/6

 

sinu = -1

 

u = 3π/2