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,

4 sin x cos x + 2 sin x + 2 cos x + 1 = 0

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

⇒ (2cos x + 1)(2sin x + 1) = 0

∴ cos x = -1/2 or sin x = -1/2

⇒ cos x = cos (π - π/3) or sin x = sin (- π/6)

⇒ cos x = cos 2π/3 or sin x = sin (-π/6)

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

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

∴ x = 2nπ ± 2π/3 or x = mπ + (-1)^m (-π/6)

Hence,