Solve the following equations :
tan x + tan 2x + tan 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,
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
∴