i know the answer is 11pi/6 & 7pi/6 & pi/2
i know sin^2 theta = (1 - cos^2 theta)
i know sin^2 theta = (1 - cos^2 theta)
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Since you already have two terms with sin, rewrite cos^2 as 1-sin^2:
Cos^2(θ)-sin^2(θ)+sin(θ)=0
1-sin^2(θ)-sin^2(θ)+sin(θ)=0
0= 2sin^2(θ)-sin(θ)-1
0= (2sinθ+1)(sinθ-1)
Sinθ=-1/2 or 1
(7pi/6, 11pi/6) or (pi/2) for θ in [0,2pi)
Hoping this helps!
Cos^2(θ)-sin^2(θ)+sin(θ)=0
1-sin^2(θ)-sin^2(θ)+sin(θ)=0
0= 2sin^2(θ)-sin(θ)-1
0= (2sinθ+1)(sinθ-1)
Sinθ=-1/2 or 1
(7pi/6, 11pi/6) or (pi/2) for θ in [0,2pi)
Hoping this helps!