Answer :


Ideas required to solve the problem:


The general solution of any trigonometric equation is given as –


• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.


• cos x = cos y, implies x = 2nπ ± y, where n Z.


• tan x = tan y, implies x = nπ + y, where n Z.


given,


5 cos2 x + 7 sin2 x – 6 = 0


5 cos2 x + 5 sin2 x + 2sin2 x – 6 = 0


2 sin2 x – 6 + 5 = 0 { sin2 x + cos2 x = 1}


2 sin2 x – 1 = 0


sin2 x = (1/2)


sin x = ±(1 /√2)


sin x = ± sin π /4


If sin x = sin y, implies x = nπ + (– 1)ny, where n Z.


x = nπ + (-1)n (±(π / 4)) where n ϵ Z


….ans


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