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Question

If 5cos2θ+2cos2θ2=1, when (0<θ<π), then the values of θ are

A
π3±π
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B
π3,cos1(35)
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C
cos1(35)±π
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D
π3,πcos1(35)
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Solution

The correct option is D π3,πcos1(35)
Given equation is
5cos2θ+2cos2θ2+1=0
5(2cos2θ1)+1+cosθ+1=0
10cos2θ+cosθ3=0
(2cosθ1)(5cosθ+3)=0
cosθ=12 or cosθ=35
θ=π3 or θ=πcos1(35)

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