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,

2 cos² x – 5 cos x + 2 =0

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

Let, cos x = k

∴ 2k² – 5k + 2 = 0

⇒ 2k² – 4k – k +2 = 0

⇒ 2k(k – 2) -1(k -2) = 0

⇒ (k – 2)(2k - 1) = 0

∴ k = 2 or k = �

⇒ cos x = 2 {which is not possible} or cos x = � (acceptable)

∴ 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