Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, 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,

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 4k² + 4k – 3 = 0

⇒ 4k² -2k + 6k – 3

⇒ 2k(2k – 1) +3(2k – 1) = 0

⇒ (2k – 1)(2k + 3) = 0

∴ k =1/2 or k = -3/2

⇒ cos x = � or cos x = -3/2

As cos x lies between -1 and 1

∴ cos x can’t be -3/2

So we ignore that value.

∴ cos x = �

⇒ cos x = cos 60° = cos π/3

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

On comparing our equation with standard form, we have

y = π/3

∴ where n ϵ Z ..ans