We will discuss about the general solution of the equation tan x minus square root of 3 equals 0 (i.e., tan x - √3 = 0) or tan x equals square root of 3 (i.e., tan x = √3).

How to find the general solution of the trigonometric equation tan x = √3 or tan x - √3 = 0?

**Solution:**

We have,

tan x - √3 = 0

⇒ tan x = √3

⇒ tan x = \(\frac{π}{3}\)

Again, tan x = √3

⇒ tan x = \(\frac{π}{3}\)

⇒ tan x = (π + \(\frac{π}{3}\))

⇒ tan x = tan \(\frac{4π}{3}\)

Let O be the centre of a unit circle. We know that in unit
circle, the length of the circumference is 2π.

If we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), and 2π.

Therefore, from the above unit circle it is clear that the final arm OP of the angle θ lies either in the first or in the final third quadrant.

If the final arm OP lies the first quadrant then,

tan x = √3

⇒ tan x = cos \(\frac{π}{3}\)

⇒ tan x = ten (2nπ + \(\frac{π}{3}\)), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Therefore, x = 2nπ + \(\frac{π}{3}\) …………….. (i)

Again, the final arm OP lies in the third quadrant then,

tan x = √3

⇒ tan x = cos \(\frac{4π}{3}\)

⇒ tan x = ten (2nπ + \(\frac{4π}{3}\)) , Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Therefore, x = 2nπ + \(\frac{π}{3}\) …………….. (ii)

Therefore, the general solution of equation tan x - √3 = 0 are the infinite sets of values of x given in (i) and (ii).

Hence general solution of tan x - √3 = 0 is **x = n****π + \(\frac{π}{3}\)**, n ∈
I.

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