Question

The general solution of $\sqrt{3}\tan \theta -1=0$ is :

• A. $\theta =n\pi +\frac{\pi }{3};n\in Z$
• B. $\theta =n\pi +\frac{\pi }{6};n\in Z$
• C. $\theta =2n\pi +\frac{\pi }{6};n\in Z$
• D. $\theta =2n\pi +\frac{\pi }{3};n\in Z$

Answer 1

The correct option is B $\theta =n\pi +\frac{\pi }{6};n\in Z$

The given equation is $\sqrt{3}\tan \theta -1=0$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$
We know that,
$\tan \theta =\tan \alpha \Rightarrow \theta =n\pi +\alpha ;n\in Z$
We also know that $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan \theta =\tan \frac{\pi }{6}$
$\Rightarrow \theta =n\pi +\frac{\pi }{6};n\in Z$