Question

Solve the following equations:
(i) $\tan x+\tan 2x+\tan 3x=0$
(ii) $\tan x+\tan 2x=\tan 3x$
(iii) $\tan 3x+\tan x=2\tan 2x$

Answer 1

(i) We have: tan x + tan 2x + tan 3x = 0
Now,
tanx+ tan2x + tan (x+ 2x) = 0

⇒tanx + tan2x + tan x+ tan 2x1 - tan x tan 2x = 0

⇒ (tanx + tan2x) (1 - tanxtan2x) + tanx + tan2x= 0

⇒(tanx + tan2x) (2 - tanx tan2x) = 0

⇒ tan x+ tan 2x= 0 or 2 - tanx tan2x = 0
Now,
tanx + tan2x = 0

⇒ tanx = - tan2x

⇒ tanx = tan -2x

⇒ x = nπ - 2x

⇒ 3x = nπ

⇒ x= nπ3, n∈Z

And,
2 - tanx tan2x = 0

⇒ tanx tan2x = 2

⇒ sinx cosx sin2x cos2x = 2

⇒ 2 sin2x cosx cosx = 2 cos2x - 2 sin2x

⇒ 4 sin2x= 2 cos2x

⇒ tan2x=12

⇒ tan2x = tan2α

⇒x = mπ + α, m ∈ Z, α = tan-1 12

∴ x= nπ3, n∈Z or x = mπ + α, m∈Z
Here,
α= tan-112
(ii) Given:
tanx+ tan2x = tan3x
Now,
tanx+ tan2x = tan (x + 2x)

⇒tanx + tan 2x = tanx + tan2x1- tanx tan2x

⇒ tanx+ tan2x - tanx + tan2x1- tanx tan2x = 0

⇒ (tanx + tan2x) (1- tanx tan2x) - (tanx + tan2x) = 0

⇒ (tanx+ tan 2x) (1- tanx tan2x -1) = 0⇒ (tanx + tan2x) (- tanx tan2x) = 0

⇒ tan x + tan 2x= 0 or tanx tan2x= 0
Now,
tan x+ tan 2x = 0

⇒ tan x = - tan 2x

⇒ tan x= tan -2x

⇒ x= nπ - 2x, n∈Z

⇒ 3x = nπ

⇒x= nπ3, n∈Z
And,
tanx tan2x = 0

⇒ sinxcosx sin2xcos2x = 0

⇒ 2 sin2xcos2x - sin2x = 0

⇒ sin2x =0

⇒ sin2x = sin20

⇒ x = mπ, m∈Z
∴ x= nπ3, n∈Z or x= mπ, m∈Z

(iii) Given: tan3x+ tanx= 2 tan2x
Now,
tan3x - tan2x= tan2x- tanx

⇒ tanx (1 + tan3x tan2x) = tanx(1 + tan2x tanx)

$\left(tan\right(A-B)=\frac{tanA-tanB}{1+tanAtanB})$

⇒tanx (1 + tan3xtan2x - 1 - tan2x tanx)= 0

⇒tanx tan2x (tan3x - tanx)= 0

⇒ tan 2x = 0 or, tan x= 0 or, tan3x- tanx = 0
And, tan 2x = 0 ⇒ 2x = nπ

⇒ x = nπ2, n∈Z
Or, tan 3x - tan x = 0

⇒ tan 3x= tan x

⇒ 3x= nπ + x

⇒ 2x=nπ

⇒ x = nπ2, n∈Z
And, tanx = 0

⇒ x = mπ, m∈Z
∴ x= nπ2, n∈Z or x = mπ, m∈Z