Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• 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.
Given,
∴ tan x = -1 or tan x = √3
As, tan x ϵ (-∞ , ∞) so both values are valid and acceptable.
⇒ tan x = tan (\(-\frac{π}4\)) or tan x = tan (\(\frac{π}3\))
If tan x = tan y, implies x = nπ + y, where n ∈ Z.
Clearly by comparing standard form with obtained equation we have
y = \(-\frac{π}4\) or y = \(\frac{π}3\)
∴ x = mπ \(-\frac{π}4\) or x = nπ + \(\frac{π}3\)
Hence,
x = mπ \(-\frac{π}4\) or nπ + \(\frac{π}3\),where m,n ϵ Z
