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