Solve the following equations :
cos x + cos 2x + cos 3x = 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,
cos x + cos 2x + cos 3x = 0
To solve the equation we need to change its form so that we can equate the t-ratios individually.
For this we will be applying transformation formulae. While applying the
Transformation formula we need to select the terms wisely which we want
to transform.
As, cos x + cos 2x + cos 3x = 0
∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.
∴ cos x + cos 2x + cos 3x = 0
⇒ cos 2x + (cos x + cos 3x) = 0
{∵ cos A + cos B = 2
⇒ cos 2x + 2 cos
⇒ cos 2x + 2cos 2x cos x = 0
⇒ cos 2x ( 1 + 2 cos x) = 0
∴ cos 2x = 0 or 1 + 2cos x = 0
⇒ cos 2x = cos π/2 or cos x = -1/2
⇒ cos 2x = cos π/2 or cos x = cos (π - π/3) = cos (2π /3)
If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.
From above expression and on comparison with standard equation we have:
y = π/2 or y = 2π/3
∴ 2x = 2nπ ± π/2 or x = 2mπ ± 2π/3
∴ where m, n ϵ Z