Given,
2sin2θ−cos2θ=0
⇒2sin2θ−(1−2sin2θ)=0
⇒sin2θ=(21)2
⇒θ=6π,65π,67π,611π
Also given, 2cos2θ+3sinθ=0
⇒2sin2θ−3sinθ−2=0
⇒sinθ=−21
⇒θ=67π,611π
So, the common solution is
θ=67π,611π
Sum =67π+11π=3π=kπ
⇒K=3
JEE Main 2022 — Mathematics Trigonometry
If the sum of solutions of the system of equations 2sin2θ−cos2θ=0 and 2cos2θ+3sinθ=0 in the interval [0,2π] is kπ, then k is equal to _______.
Held on 26 Jul 2022 · Verified 6 Jul 2026.
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