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

$\begin{array}{1 1}(A)\;\large\frac{\pi}{3},\frac{4\pi}{3},\normalsize x=n\pi+\large\frac{\pi}{3}\\(B)\;\large\frac{5\pi}{6},\frac{11\pi}{6},\normalsize x=n\pi+\large\frac{5\pi}{6}\\(C)\;\large\frac{\pi}{6},\frac{7\pi}{6},\normalsize x=2n\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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  • $\cot \large\frac{\pi}{6}$$=\sqrt 3,\tan \large\frac{\pi}{6}=\frac{1}{\sqrt 3}$
  • $\tan x=\tan y,x=n\pi+y$
$\cot \large\frac{\pi}{6}$$=\sqrt 3,\tan \large\frac{\pi}{6}=\frac{1}{\sqrt 3}$
$\tan(\pi-\large\frac{\pi}{6})=$$-\tan \large\frac{\pi}{6}=$$\tan \large\frac{5\pi}{6}=-\large\frac{1}{\sqrt 3}$
$\tan(2\pi-\large\frac{\pi}{6})$$=-\tan \large\frac{\pi}{3}=-\frac{1}{\sqrt 3}$
$\therefore \tan \large\frac{5\pi}{6}$$=\tan \large\frac{11\pi}{6}=-\large\frac{1}{\sqrt 3}$
$\therefore \cot \large\frac{5\pi}{6}$$=\cot \large\frac{11\pi}{6}$$=-\sqrt 3$
Principal solutions are $\large\frac{5\pi}{6},\frac{11\pi}{6}$
$\cot x=\cot \large\frac{5\pi}{6}$
General solution :$x=n\pi+\large\frac{5\pi}{6}$ where $n\in z$
Hence (B) is the correct answer.
answered Apr 17, 2014 by sreemathi.v
 

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