Solve the equation. Cos^2θ-sin^2θ=1+sinθ on the interval 0<θ<2π

Question

Please help! This is for my precal class test review and I’ve been trying to solve it for an hour now and can’t seem to figure out how to do it. Here is the problem once again:

cos²θ-sin²θ=1+sinθ on the interval 0≤θ<2π

Answer 1

let θ=x for simplicity on my computer

cos^2x-sin^2x=1+sinx

sin^2x+cos^2x=1 is a trig identity

therefore cos^2x=1-sin^2x

substitute

1-sin^2x-sin^2x=1+sinx

1-2sin^2x=1+sinx

subtract 1 from both sides

-2sin^2x=sinx

2sin^2x+sinx=0

factor

sinx(2sinx+1)=0

set each factor equal to 0

sinx=0

look on the unit circle

sin is 0 when x=0º and x=180º

2sinx+1=0

2sinx=-1

sinx=(-1/2)

look on the unit circle

sin is (-1/2) when x=7∏/6 or 210° and x=11∏/6 or 330º