Trigonometric Equations Question 37
Question: If $ \sec x\cos 5x+1=0 $ , where $ 0<x<2\pi $ , then x =
[Roorkee 1978; IIT 1963]
Options:
A) $ \frac{\pi }{5},\frac{\pi }{5} $
B) $ \frac{\pi }{5} $
C) $ \frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \sec x\cos 5x=-1\Rightarrow \cos 5x=-\cos x $
$ \Rightarrow $ $ 5x=2n\pi \pm (\pi -x) $
$ \Rightarrow $ $ x=\frac{(2n+1)\pi }{6}\text{ or }\frac{(2n-1)\pi }{4} $ Hence $ x=\frac{\pi }{4},,\frac{\pi }{2},,\frac{3\pi }{4},,\frac{5\pi }{6},,\frac{5\pi }{4},,\frac{7\pi }{6},,\frac{7\pi }{4},,\frac{9\pi }{6},,\frac{11\pi }{6} $ .
BETA
