Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, 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.

(i) Given as cos x + cos 2x + cos 3x = 0

Now let us simplify,

cos x + cos 2x + cos 3x = 0

Then, we shall rearrange and use transformation formula

cos 2x + (cos x + cos 3x) = 0

On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2

cos 2x + 2 cos (3x + x)/2 cos (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 0 or cos x = -1/2

cos 2x = cos π/2 or cos x = cos (π – π/3)

cos 2x = cos π/2 or cos x = cos (2π/3)

2x = (2n + 1) π/2 or x = 2mπ ± 2π/3

x = (2n + 1) π/4 or x = 2mπ ± 2π/3

∴ the general solution is

x = (2n + 1) π/4 or 2mπ ± 2π/3, where m, n ϵ Z.

(ii) cos x + cos 3x – cos 2x = 0

Now let us simplify,

cos x + cos 3x – cos 2x = 0

Then, we shall rearrange and use transformation formula

cos x – cos 2x + cos 3x = 0

– cos 2x + (cos x + cos 3x) = 0

On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2

– cos 2x + 2 cos (3x + x)/2 cos (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 0 or cos x = 1/2

cos 2x = cos π/2 or cos x = cos (π/3)

2x = (2n + 1) π/2 or x = 2mπ ± π/3

x = (2n + 1) π/4 or x = 2mπ ± π/3

∴ the general solution is

x = (2n + 1) π/4 or 2mπ ± π/3, where m, n ϵ Z.

(iii) sin x + sin 5x = sin 3x

Now let us simplify,

sin x + sin 5x = sin 3x

sin x + sin 5x – sin 3x = 0

Then, we shall rearrange and use transformation formula

– sin 3x + sin x + sin 5x = 0

– sin 3x + (sin 5x + sin x) = 0

On using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2

– sin 3x + 2 sin (5x + x)/2 cos (5x - x)/2 = 0

2sin 3x cos 2x – sin 3x = 0

sin 3x ( 2cos 2x – 1) = 0

sin 3x = 0 or 2cos 2x – 1 = 0

sin 3x = sin 0 or cos 2x = 1/2

sin 3x = sin 0 or cos 2x = cos π/3

3x = nπ or 2x = 2mπ ± π/3

x = nπ/3 or x = mπ ± π/6

∴ the general solution is

x = nπ/3 or mπ ± π/6, where m, n ϵ Z.

(iv) Given as cos x cos 2x cos 3x = 1/4

Now let us simplify,

cos x cos 2x cos 3x = 1/4

4 cos x cos 2x cos 3x – 1 = 0

On using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2(2cos x cos 3x) cos 2x – 1 = 0

2(cos 4x + cos 2x) cos2x – 1 = 0

2(2cos² 2x – 1 + cos 2x) cos 2x – 1 = 0 [using cos 2A = 2cos²A – 1]

4cos³ 2x – 2cos 2x + 2cos² 2x – 1 = 0

2cos² 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

(2cos² 2x – 1) (2 cos 2x + 1) = 0

Therefore,
2cos 2x + 1 = 0 or (2cos² 2x – 1) = 0

cos 2x = -1/2 or cos 4x = 0 [using cos 2θ = 2cos²θ – 1]

cos 2x = cos (π – π/3) or cos 4x = cos π/2

cos 2x = cos 2π/3 or cos 4x = cos π/2

2x = 2mπ ± 2π/3 or 4x = (2n + 1) π/2

x = mπ ± π/3 or x = (2n + 1) π/8

∴ the general solution is

Thus, x = mπ ± π/3 or (2n + 1) π/8, where m, n ϵ Z.