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 = nπ + \(\frac{π}{3}\), n ∈ I.
`11 and 12 Grade Math
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