Question

The general solution of tan3x = 1 is
$
  (a){\text{ n}}\pi {\text{ + }}\dfrac{\pi }{4} \\
  (b){\text{ }}\dfrac{{n\pi }}{3}{\text{ + }}\dfrac{\pi }{{12}} \\
  (c){\text{ n}}\pi \\
  (d){\text{ n}}\pi \pm \dfrac{\pi }{4} \\
 $

Answer 1

Hint – In this question use the concept that $\tan \theta $ is positive in first and third quadrant hence use ${\text{tan}}(\pi + \theta ) = \tan \theta $ along with $\tan \dfrac{\pi }{4} = 1$, to find the general solution use the generalization of the above concept to infer ${\text{tan}}(n\pi + \theta ) = \tan \theta $.

Complete step-by-step answer:

Given trigonometric equation
$\tan 3x = 1$........................ (1)
Now as we know that $\tan \dfrac{\pi }{4} = 1$
Now we all know tan is positive in first and third quadrant and ${\text{tan}}(\pi + \theta ) = \tan \theta $
So $\tan \left( {\pi + \dfrac{\pi }{4}} \right) = \tan \dfrac{\pi }{4} = 1$
So in general we can say that
$ \Rightarrow \tan \left( {n\pi + \dfrac{\pi }{4}} \right) = \tan \dfrac{\pi }{4} = 1$............................ (2)
Therefore from equation (1) and (2) we have,
$ \Rightarrow \tan 3x = \tan \left( {n\pi + \dfrac{\pi }{4}} \right)$
$ \Rightarrow 3x = n\pi + \dfrac{\pi }{4}$
Now divide by 3 we have,
$ \Rightarrow x = \dfrac{{n\pi }}{3} + \dfrac{\pi }{{12}}$
So this is the required general solution.
Hence option (B) is correct.

Note – The periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. Thus obtaining all these solutions in terms of a variable n such that n can belong to a particular set of numbers, is the basic idea behind the general solution in trigonometric equations.