Trigonometric Equations Question 355
Question: If $ {{\tan }^{2}}\theta -(1+\sqrt{3})\tan \theta +\sqrt{3}=0 $ , then the general value of $ \theta $ is
Options:
A) $ n\pi +\frac{\pi }{4},n\pi +\frac{\pi }{3} $
B) $ n\pi -\frac{\pi }{4},n\pi +\frac{\pi }{3} $
C) $ n\pi +\frac{\pi }{4},n\pi -\frac{\pi }{3} $
D) $ n\pi -\frac{\pi }{4},n\pi -\frac{\pi }{3} $
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Answer:
Correct Answer: A
Solution:
- $ {{\tan }^{2}}\theta -\tan \theta -\sqrt{3}\tan \theta +\sqrt{3}=0 $
$ \Rightarrow $ $ \tan \theta (\tan \theta -1)-\sqrt{3}(\tan \theta -1)=0 $
$ \Rightarrow $ $ (\tan \theta -\sqrt{3})(\tan \theta -1)=0 $
$ \Rightarrow $ $ \theta =n\pi +\frac{\pi }{3} $ , $ n\pi +\frac{\pi }{4} $ .
BETA
