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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the principal and general solution of the following equation $\tan x=\sqrt 3$

$\begin{array}{1 1}(A)\;\large\frac{\pi}{3},\frac{4\pi}{3},\normalsize x=n\pi+\large\frac{\pi}{3}\\(B)\;\large\frac{7\pi}{3},\frac{2\pi}{3},\normalsize x=n\pi+\large\frac{\pi}{2}\\(C)\;\large\frac{\pi}{5},\frac{4\pi}{5},\normalsize x=n\pi+\large\frac{\pi}{5}\\(D)\;\large\frac{\pi}{7},\frac{4\pi}{7},\normalsize x=n\pi+\large\frac{\pi}{7}\end{array} $

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1 Answer

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Toolbox:
  • $\tan \large\frac{\pi}{3}$$=\sqrt 3$
  • $\tan x=\tan y$ when $y$ is the principal value of $x$
  • $x=n\pi+y$
We know that $\tan \large\frac{\pi}{3}$$=\sqrt 3$
$\tan (\pi+\large\frac{\pi}{3})=$$\tan \large\frac{\pi}{3}$$=\sqrt 3$
$\Rightarrow \tan \large\frac{4\pi}{3}$
$\therefore \tan \large\frac{\pi}{3}=$$\tan \large\frac{4\pi}{3}$$=\sqrt 3$
$\therefore$ Principal solutions are $\large\frac{\pi}{3},\frac{4\pi}{3}$
If $\tan x=\tan \large\frac{\pi}{3}$
$\therefore$ General solution: $x=n\pi+\large\frac{\pi}{3}$where $n \in z$
Hence (A) is the correct answer.
answered Apr 17, 2014 by sreemathi.v
 

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