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,

2 sin² x = 3 cos x , 0 ≤ x ≤ 2π

⇒ 2 ( 1 – cos² x) = 3 cos x

⇒ 2 cos² x + 3cos x – 2 = 0

⇒ 2 cos² x + 4 cos x – cos x – 2 = 0

⇒ 2 cos x(cos x + 2) – 1(cos x + 2) = 0

⇒ (2cos x – 1)(cos x + 2) = 0

∴ cos x = �

or cos x = -2 { as cos x lies between -1 and 1 so this value is rejected }

∴ cos x = � = cos π/3

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

∴ x = 2nπ ± π/3

But, 0 ≤ x ≤ 2π

∴ x = π/3 and x = 2π - π/3 = 5π/3 ….ans