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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