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Find the general solution of the following equation: 

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

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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

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