Q.
If 6cos2θ+2cos2θ/2+2sin2θ=0 where −π<0<π , then θ is equal to
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a
π/3
b
π/3,cos−13/5
c
cos−13/5
d
π/3,π−cos−13/5
answer is D.
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Detailed Solution
The given equation can be written as62cos2θ−1+1+cosθ+21−cos2θ=0⇒10cos2θ+cosθ−3=0⇒5cosθ+32cosθ−1=0⇒cosθ=12 or ⇒cosθ=-35⇒θ=π3 or θ=π−cos−135 as −π<θ<π⇒θ=π3,π−cos−135
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