Trigonometric Equations Question 10
Question: The general value of $ \theta $ in the equation $ 2\sqrt{3}\cos \theta =\tan \theta $ , is
[MP PET 2003]
Options:
A) $ 2n\pi \pm \frac{\pi }{6} $
B) $ 2n\pi \pm \frac{\pi }{4} $
C) $ n\pi +{{(-1)}^{n}}\frac{\pi }{3} $
D) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4} $
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Answer:
Correct Answer: C
Solution:
- $ 2\sqrt{3}{{\cos }^{2}}\theta =\sin \theta $
Þ $ 2\sqrt{3}{{\sin }^{2}}\theta +\sin \theta -2\sqrt{3}=0 $
Þ $ \sin \theta =\frac{-1\pm 7}{4\sqrt{3}}\Rightarrow \sin \theta =\frac{-8}{4\sqrt{3}}\text{,} $ (Impossible) and $ \sin \theta =\frac{6}{4\sqrt{3}}=\frac{\sqrt{3}}{2} $ Þ $ \theta =n\pi +{{(-1)}^{n}}\frac{\pi }{3} $ .
BETA
