Find the general solution of the following equation: 4sin x cos x +

1.	Find the general solution of the following equation: 4sin x cos x + 2sin x + 2cos x + 1 = 0
Answer» 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

Find the general solution of the following equation: 4sin x cos x + 2sin x + 2cos x + 1 = 0

Answer»

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