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) tan x + tan 2x + tan 3x = 0

Now let us simplify,

tan x + tan 2x + tan 3x = 0

tan x + tan 2x + tan (x + 2x) = 0

On using the formula,

tan (A + B) = [tan A + tan B]/[1 – tan A tan B]

Therefore,

tan x + tan 2x + [[tan x + tan 2x]/[1- tan x tan 2x]] = 0

(tan x + tan 2x) (1 + 1/(1- tan x tan 2x)) = 0

(tan x + tan 2x) ([2 – tan x tan 2x]/[1 – tan x tan 2x]) = 0

Now,

(tan x + tan 2x) = 0 or ([2 – tan x tan 2x]/[1 – tan x tan 2x]) = 0

(tan x + tan 2x) = 0 or [2 – tan x tan 2x] = 0

tan x = tan (-2x) or tan x tan 2x = 2

x = nπ + (-2x) or tax x [2tan x/(1 – tan² x)] = 2 [Using, tan 2x = 2 tan x/1 - tan² x]

3x = nπ or 2 tan² x/(1 - tan² x) = 2

3x = nπ or 2 tan² x = 2(1 – tan² x)

3x = nπ or 2 tan² x = 2 – 2tan² x

3x = nπ or 4 tan² x = 2

x = nπ/3 or tan² x = 2/4

x = nπ/3 or tan² x = 1/2

x = nπ/3 or tan x = 1/√2

x = nπ/3 or x = tan α [let 1/√2 be ‘α’]

x = nπ/3 or x = mπ + α

∴ the general solution is

Thus, x = nπ/3 or mπ + α, where α = tan^-11/√2, m, n ∈ Z.

(ii) tan x + tan 2x = tan 3x

Now let us simplify,

tan x + tan 2x = tan 3x

tan x + tan 2x – tan 3x = 0

tan x + tan 2x – tan (x + 2x) = 0

On using the formula,

tan (A+B) = [tan A + tan B] / [1 – tan A tan B]

Therefore,

tan x + tan 2x – [[tan x + tan 2x]/[1- tan x tan 2x]] = 0

(tan x + tan 2x) (1 – 1/(1- tan x tan 2x)) = 0

(tan x + tan 2x) ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0

Now,

(tan x + tan 2x) = 0 or ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0

(tan x + tan 2x) = 0 or [– tan x tan 2x] = 0

tan x = tan (-2x) or -tan x tan 2x = 0

tan x = tan (-2x) or 2tan² x / (1 – tan² x) = 0 [Using, tan 2x = 2 tan x / 1-tan² x]

x = nπ + (-2x) or x = mπ + 0

3x = nπ or x = mπ

x = nπ/3 or x = mπ

∴ the general solution is

x = nπ/3 or mπ, where m, n ∈ Z.

(iii) tan 3x + tan x = 2 tan 2x

Now let us simplify,

tan 3x + tan x = 2 tan 2x

tan 3x + tan x = tan 2x + tan 2x

Then, upon rearranging we get,

tan 3x – tan 2x = tan 2x – tan x

On using the formula,

tan (A - B) = [tan A – tan B]/[1 + tan A tan B]

Therefore,

[(tan 3x – tan 2x) (1 + tan 3x tan 2x)]/[1 + tan 3x tan 2x] = [(tan 2x - tan x) (1 + tan x tan 2x)]/[1 + tan 2x tan x]

tan (3x – 2x) (1 + tan 3x tan 2x) = tan (2x – x) (1 + tan x tan 2x)

tan x [1 + tan 3x tan 2x – 1 – tan 2x tan x] = 0

tan x tan 2x (tan 3x – tan x) = 0

Therefore,

tan x = 0 or tan 2x = 0 or (tan 3x – tan x) = 0

tan x = 0 or tan 2x = 0 or tan 3x = tan x

x = nπ or 2x = mπ or 3x = kπ + x

x = nπ or x = mπ/2 or 2x = kπ

x = nπ or x = mπ/2 or x = kπ/2

∴ the general solution is

Thus, x = nπ or mπ/2 or kπ/2, where, m, n, k ∈ Z.