Q. 185.0( 5 Votes )

# If 2sin2</su

Answer :

Given,

2sin2θ = 3cos θ

We know that – sin2θ = 1 – cos2θ 2(1 – cos2θ) = 3cos θ

2 – 2cos2θ = 3cos θ

2cos2θ + 3cos θ - 2 = 0

2cos2θ + 4cos θ - cos θ - 2 = 0

2cos θ (cos θ+ 2) +1 (cos θ + 2) = 0

(2cos θ + 1)(cos θ + 2) = 0

cos θ [-1,1] , for any value θ.

So, cos θ ≠ - 2

2 cos θ - 1 = 0

cos θ = 1/2  Rate this question :

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