# Find the general

Given,

5cos2θ + 7sin2θ – 6 = 0

We know that : sin2θ = 1 – cos2θ

5cos2θ + 7(1 – cos2θ) – 6 = 0

5cos2θ + 7 – 7cos2θ – 6 = 0

-2cos2θ + 1 = 0

cos2θ = 1/2

cos θ = ±1√2

cos θ = cos π/4 or cos θ = cos 3π/4

solution of cos x = cos α is given by

x = 2mπ ± α m Z

θ = nπ ± π/4, n Z

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