Given :
[tex]tan \ A = \dfrac{1}{\sqrt{3}}[/tex]
To Find :
The general solution of the given equation.
Solution :
We know, [tex]tan\ 30^o = \dfrac{1}{\sqrt{3}}[/tex]
Comparing it with given equation, we get :
[tex]A = \dfrac{\pi}{6}[/tex]
General solution is given by :
[tex]A = n\pi + \dfrac{\pi}{6}[/tex] where n ∈ Z ( integers )
Hence, this is the required solution.
