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

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

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1 Answer

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  • $\sec\large\frac{\pi}{3}=\frac{1}{\cos\Large\frac{\pi}{3}}$$=2$
  • $\cos x=\cos y,x=2n\pi\pm y$
We know that $\cos\large\frac{\pi}{3}=\frac{1}{2}$
$\therefore \sec\large\frac{\pi}{3}$$=2$
$\sec(2\pi-\large\frac{\pi}{3})$$=\sec\large\frac{5\pi}{3}$$=2$
$\therefore$ Principal solution are $\large\frac{\pi}{3},\frac{5\pi}{3}$
If $\sec x=\sec\large\frac{\pi}{3}$
$\therefore$ General solution $x=2n\pi\pm \large\frac{\pi}{3}$where $n\in z$
Hence (A) is the correct answer.
answered Apr 17, 2014 by sreemathi.v
 

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