Q. 5 A4.7( 3 Votes )

# 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

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