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

2sin2xsinx-10

#1

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