Q.

The sum of all values of θ∈[0,2π] satisfying 2sin2θ=cos2θ and 2cos2θ=3sinθ is:          [2025]

1 π/2  
2 π  
3 5π/6  
4 4π  

Ans.

(2)

2sin2θ=cos2θ

⇒ 2sin2θ=1–2sin2θ ⇒ 4sin2θ=1

⇒ sin2θ=14 ⇒ sinθ=±12          ... (i)

Also, 2cos2θ=3sinθ

⇒ 2(1–sin2θ)=3sinθ

⇒ 2sin2θ+3sinθ–2=0

⇒ 2sin2θ+4sinθ–sinθ–2=0

⇒ (2sinθ–1)(sinθ+2)=0

⇒ sinθ=12(sinθ=–2(not possible))          ... (ii)

Thus, θ=π6,5π6(θ∈[0,2π])          [Using (i) & (ii)]

∴  Required sum = π