To Find: General solution.
Given: 4sin x cos x + 2sin x + 2cos x + 1 = 0
⇒ 2sin x (2cos x + 1) + 2cos x + 1 = 0
So (2cos x + 1) (2sin x + 1) = 0
cos x -1/2 = cos(2π)3 or x = -1/2 = sin 7π/6
Formula used: cos θ = cos α
θ = 2nπ ± α or sin θ = sin α
⇒ θ = mπ + (-1)m α where n, m ∈ l
x = 2nπ ± 2π/3 or x = mπ + (-1)m. 7π/6 where n, m ∈ l
So the general solution is x = 2nπ ± 2π/3 or x = mπ + (-1)m. 7π/6 where n, m ∈ l