If 5cos2θ+2cos2θ2=−1, when (0<θ<π), then the values of θ are
A
π3±π
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B
π3,cos−1(35)
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C
cos−1(35)±π
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D
π3,π−cos−1(35)
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Solution
The correct option is Dπ3,π−cos−1(35) Given equation is 5cos2θ+2cos2θ2+1=0 ⇒5(2cos2θ−1)+1+cosθ+1=0 ⇒10cos2θ+cosθ−3=0 ⇒(2cosθ−1)(5cosθ+3)=0 ⇒cosθ=12 or cosθ=−35 ⇒θ=π3 or θ=π−cos−1(35)