Solve the following equations :
4 sin x cos x + 2 sin x + 2 cos x + 1 = 0
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,