Question -
Answer -
The general solution of any trigonometric equation is givenas:
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
Let us simplify,
tan x + tan 2x + tan 3x = 0
tan x + tan 2x + tan (x + 2x) = 0
By using the formula,
tan (A+B) = [tan A + tan B] / [1 – tan A tan B]
So,
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
Then,
(tan x + tan 2x) = 0 or ([2 – tan x tan 2x] / [1 – tan x tan2x]) = 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
x = nπ/3 or mπ + α, where α = tan-11/√2, m, n ∈ Z.
(ii) tan x + tan 2x = tan 3x
Let us simplify,
tan x + tan 2x = tan 3x
tan x + tan 2x – tan 3x = 0
tan x + tan 2x – tan (x + 2x) = 0
By using the formula,
tan (A+B) = [tan A + tan B] / [1 – tan A tan B]
So,
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
Then,
(tan x + tan 2x) = 0 or ([– tan x tan 2x] / [1 – tan x tan2x]) = 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
Let us simplify,
tan 3x + tan x = 2 tan 2x
tan 3x + tan x = tan 2x + tan 2x
upon rearranging we get,
tan 3x – tan 2x = tan 2x – tan x
By using the formula,
tan (A-B) = [tan A – tan B] / [1 + tan A tan B]
so,
[(tan 3x – tan 2x) (1+tan 3x tan2x)] / [1 + tan 3x tan 2x] = [(tan 2x-tan x) (1+tan x tan 2x)] / [1 + tan 2xtan x]
tan (3x – 2x) (1 + tan 3x tan 2x) = tan (2x – x) (1 + tan xtan 2x)
tan x [1 + tan 3x tan 2x – 1 – tan 2x tan x] = 0
tan x tan 2x (tan 3x – tan x) = 0
so,
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
x = nπ or mπ/2 or kπ/2, where, m, n, k ∈ Z.