# Find the value of

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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