If θ ∈ [0, 2π] satisfying the system of equations 2sin^2θ and 2cos^2θ = 3sinθ. Then the sum of all real values of θ is

Question

If \(\theta \in[0,2 \pi]\) satisfying the system of equations \(2 \sin ^2 \theta=\cos 2 \theta\) and \(2 \cos ^2 \theta=3 \sin \theta.\) Then the sum of all real values of \(\theta\) is

(1) \(\frac{3 \pi}{2}\)

(2) \(\pi\)

(3) \(\frac{\pi}{2}\)

(4) \(\frac{5 \pi}{6}\)

Answer 1

Correct option is (2) \(\pi\)

\(2 \sin ^2 \theta=\cos 2 \theta\)  

\(2 \cos ^2 \theta=3 \sin \theta\)

\(\Rightarrow\) Adding, 

\(2= 1-2 \sin ^2 \theta+3 \sin \theta \) 

\(\Rightarrow 2 \sin ^2 \theta-3 \sin \theta+1=0 \)  

\(2 \sin ^2 \theta-2 \sin \theta-\sin \theta+1=0 \)  

\(2 \sin \theta(\sin \theta-1)-1(\sin \theta-1)=0 \)    

\(\sin \theta=1, \frac{1}{2} \text { but } 2 \sin ^2 \theta=\cos 2 \theta=2​​​​​​​\)   

but not is not possible   

\(\Rightarrow \theta=\frac{\pi}{6},\left(\pi-\frac{\pi}{6}\right)\)  

\(\Rightarrow\) Sum of all values \(=\frac{\pi}{6}+\pi-\frac{\pi}{6}=\pi\)