Q. 124.7( 6 Votes )

Find the value of

Answer :

We have,

Expanding along R1, we get




(1){-6 – {(-7) cos2θ}} – 1{8 – 7cos2θ} + sin3θ {28 – 21} = 0


– 6 + 7cos2θ – 8 + 7cos2θ + 7sin3θ = 0


14cos2θ + 7sin3θ – 14 = 0


2cos2θ + sin3θ – 2 = 0


Now, we know that


cos 2θ = 1 – 2sin2θ


sin 3θ = 3sinθ – 4sin3θ


2(1 – 2sin2θ) + (3sinθ – 4sin3θ) – 2 = 0


2 – 4sin2θ + 3sinθ – 4sin3θ – 2 = 0


-2 + 4sin2θ - 3sinθ + 4sin3θ + 2 = 0


sinθ (4sinθ – 3 + 4sin2θ) = 0


sinθ (4sin2θ – 6sinθ + 2sinθ – 3) = 0


sinθ [2sinθ(2sinθ – 3) + 1(2sinθ – 3)] = 0


sinθ (2sinθ + 1)(2sinθ – 3) = 0


sinθ = 0 or 2sinθ + 1 = 0 or 2sinθ – 3 = 0


θ = nπ or 2sinθ = -1 or 2sinθ = 3



θ = nπ ; m, n Z



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