SOLUTION: I need to solve this equation for the interval [0, 2pi) : -sin^2(x) = -2sin^2(x)-sinx

Algebra ->  Trigonometry-basics -> SOLUTION: I need to solve this equation for the interval [0, 2pi) : -sin^2(x) = -2sin^2(x)-sinx      Log On


   

Question 1063373: I need to solve this equation for the interval [0, 2pi) :
-sin^2(x) = -2sin^2(x)-sinx
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
-sin^2(x) = -2sin^2(x)-sinx
sin^2 x=-sin x
sin^2 x+ sin x=0
sin x(sin x+1)=0
sin x=0, -1
At 0, pi, and 3pi/2