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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the principal and general solution of the following equation $cosec x=-2$

$\begin{array}{1 1}(A)\;\large\frac{7\pi}{6},\frac{11\pi}{3},\normalsize x=n\pi+(-1)^n\large\frac{7\pi}{6}\\(B)\;\large\frac{7\pi}{3},\frac{2\pi}{3},\normalsize x=n\pi+(-1)^n\large\frac{\pi}{2}\\(C)\;\large\frac{\pi}{5},\frac{4\pi}{5},\normalsize x=n\pi+(-1)^n\large\frac{\pi}{5}\\(D)\;\large\frac{\pi}{7},\frac{4\pi}{7},\normalsize x=n\pi+(-1)^n\large\frac{\pi}{7}\end{array} $

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1 Answer

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  • $cosec \large\frac{\pi}{6}$$=2,\sin \large\frac{\pi}{6}=\frac{1}{2}$
  • $\sin x=\sin y$
  • $\therefore x=n\pi+(-1)^ny$
$cosec (\pi+\large\frac{\pi}{6})$$=-cosec \large\frac{\pi}{6}=$$-2$
$\sin (\pi+\large\frac{\pi}{6})$$=-\sin \large\frac{\pi}{6}=-\frac{1}{2}$
$cosec \large\frac{7\pi}{6}=$$-2$
$\sin (2\pi-\large\frac{\pi}{6})=$$-\sin \large\frac{\pi}{6}=-\frac{1}{2}$
$\Rightarrow \sin \large\frac{11}{6}$$\pi=-\large\frac{1}{2}$
$cosec \large\frac{11\pi}{6}$$=-2$
Principal solutions are $\large\frac{7\pi}{6},\frac{11\pi}{6}$
$\therefore cosec x=cosec \large\frac{7\pi}{6}$
General solution $x=n\pi+(-1)^n\large\frac{7\pi}{6}$ where $n \in z$
Hence (A) is the correct answer.
answered Apr 17, 2014 by sreemathi.v
 

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