Trigonometric Equations Question 341
Question: If $ \sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1 $ , then the general value of $ \theta $ is
Options:
A) $ n\pi +\frac{\pi }{5} $
B) $ ( n+\frac{1}{6} )\frac{\pi }{5} $
C) $ ( 2n\pm \frac{1}{6} )\frac{\pi }{5} $
D) $ ( n+\frac{1}{3} )\frac{\pi }{5} $
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Answer:
Correct Answer: B
Solution:
- $ \sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1 $
$ \Rightarrow $ $ \frac{\tan 2\theta +\tan 3\theta }{1-\tan 2\theta \tan 3\theta }=\frac{1}{\sqrt{3}} $
Þ $ \tan 5\theta =\tan \frac{\pi }{6} $
$ \Rightarrow $ $ 5\theta =n\pi +\frac{\pi }{6}\Rightarrow \theta =( n+\frac{1}{6} )\frac{\pi }{5} $ .
BETA
