Trigonometric Equations Question 335
Question: If $ \sqrt{3}\cos ,\theta +\sin \theta =\sqrt{2}, $ then the most general value of $ \theta $ is
[MP PET 1991, 2002; UPSEAT 1999]
Options:
A) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4} $
B) $ {{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3} $
C) $ n\pi +\frac{\pi }{4}-\frac{\pi }{3} $
D) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3} $
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Answer:
Correct Answer: D
Solution:
- $ \frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta =\frac{\sqrt{2}}{2} $ {dividing by $ \sqrt{{{(\sqrt{3})}^{2}}+1^{2}}=2} $
$ \Rightarrow $ $ \sin ( \theta +\frac{\pi }{3} )=\frac{1}{\sqrt{2}}=\sin ( \frac{\pi }{4} ) $
$ \Rightarrow $ $ \theta =n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3} $ .
BETA
