**find all solutions of each equation on the interval [0,2π) 2sin^2x+sinx-1=0**

**Answer:**

Well, you can go ahead and set the entire thing equal to 1.

2sin^2x+sinx=1

factor out sinx from left side

sinx(2sinx+1)=1

and set each of the factors equal to one and solve

sinx=1 and 2sinx+1=1

x=pi/2