Solve for theta on the interval [0, 2pi): 2sin^2x -sin x=0

Question

Don't understand please help

Answer 1

We can factor the left side of the equation by a common factor.

 

sinx(2sinx - 1) = 0

 

 

Set the factors equal to zero.

 

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

 

x = 0                     and                    sin(x) = 1 / 2

 

x = 2π                                                  x = π / 6

                        

                                                         

sine is positive in the 1st and 2nd quadrants.

 

x = π - (π / 6) = 5π / 6

 

 

We have 4 solutions:

 

x = 0

x = π / 6

x = 5π / 6

x = 2π