Solve the followi

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,

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

In order to solve the equation we need to reduce the equation into factor form so that we can equate the ratios with 0 and can solve the equation easily

As if we expand tan 3x = tan ( x + 2x) we will get tan x + tan 2x common.

∴ tan x + tan 2x + tan 3x = 0

⇒ tan x + tan 2x + tan (x + 2x) = 0

As, tan (A + B) =

∴ tan x + tan 2x +

⇒

⇒

∴ tan x + tan 2x = 0 or 2 – tan x tan 2x = 0

Using, tan 2x = we have,

⇒ tan x = tan (-2x) or 2 –

⇒ tan x = tan(-2x) or 2 – 4tan² x = 0 ⇒ tan x = 1/ √2

Let 1/√2 = tan α and if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ + (-2x) or tan x = tan α ⇒ x = mπ + α

⇒ 3x = nπ or x = mπ + α

∴