Trigonometric Equations Question 383
Question: The general solution of the equation $ (\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2 $ is
[Roorkee 1992]
Options:
A) $ 2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12} $
B) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{12} $
C) $ 2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12} $
D) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{12} $
Show Answer
Answer:
Correct Answer: A
Solution:
- Let $ \sqrt{3}+1=r\cos \alpha $ and $ \sqrt{3}-1=r\sin \alpha $ . Then $ r=\sqrt{{{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}}=2\sqrt{2} $ tan $ \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{1-(1/\sqrt{3})}{1+(1/\sqrt{3})}=\tan ( \frac{\pi }{4}-\frac{\pi }{6} )\Rightarrow \alpha =\frac{\pi }{12} $ The given equation reduces to $ 2\sqrt{2}\cos (\theta -\alpha )=2\Rightarrow \cos ( \theta -\frac{\pi }{12} )=\cos \frac{\pi }{4} $
$ \Rightarrow $ $ \theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4}\Rightarrow \theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12} $ .
BETA
