Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • $\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$
$\therefore$ 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App