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 \((\frac{A+B}2)\) cos \((\frac{A-B}2)\)]

⇒ cos 2x + 2 cos \((\frac{3x+x}2)\) cos \(\frac{3x-x}2\) = 0

⇒ 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 = \(\frac{π}2\) or y = \(\frac{2π}3\)

∴ 2x = 2nπ ± \(\frac{π}2\) or x = 2mπ ± \(\frac{2π}3\)

∴ x = nπ ± \(\frac{π}4\) or x = 2mπ ± \(\frac{2π}3\) where m, n ϵ Z