Answer :

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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