+561 \(sin2\theta=cos\theta \)
Apply the double angle identity to sin2θ.
\(2sin\theta cos\theta=cos\theta \)
\(2sin\theta cos\theta-cos\theta=0\)
\((cos\theta)(2sin\theta-1)=0\)
\(cos\theta=0\)
\(\theta=90°,270°\)
\(2sin\theta-1=0\)
\(\frac{1}{2}=sin\theta\)
\(\theta=30°,150°\)
Therefore, θ can be equal to 30°, 90°, 150°, 270°
+561 \(sin2\theta=cos\theta \)
Apply the double angle identity to sin2θ.
\(2sin\theta cos\theta=cos\theta \)
\(2sin\theta cos\theta-cos\theta=0\)
\((cos\theta)(2sin\theta-1)=0\)
\(cos\theta=0\)
\(\theta=90°,270°\)
\(2sin\theta-1=0\)
\(\frac{1}{2}=sin\theta\)
\(\theta=30°,150°\)
Therefore, θ can be equal to 30°, 90°, 150°, 270°
THANK YOU 
