Trigonometric Identities Question 216
Question: The least positive solution of $ \cot ( \frac{\pi }{3\sqrt{3}}\sin 2x )=\sqrt{3} $ lies in
Options:
A) $ ( 0,\frac{\pi }{6} ] $
B) $ ( \frac{\pi }{9},\frac{\pi }{6} ) $
C) $ ( \frac{\pi }{12},\frac{\pi }{9} ] $
D) $ ( \frac{\pi }{3},\frac{\pi }{2} ] $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \cot ( \frac{\pi }{3\sqrt{3}}\sin 2x )=\sqrt{3} $
$ \Rightarrow ,\frac{\pi }{3\sqrt{3}}\sin 2x=n\pi +\frac{\pi }{6},n\in Z $
$ \Rightarrow ,\sin 2x=3\sqrt{3}n+\frac{\sqrt{3}}{2} $ For least positive solution, n=0
$ \Rightarrow \sin 2x=\frac{\sqrt{3}}{2} $
$ \Rightarrow 2x=\frac{\pi }{3}\Rightarrow x=\frac{\pi }{6} $
BETA
