Trigonometric Identities Question 260
Question: General solution of the equation $ (\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2 $ is
Options:
A) $ 2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12} $
B) $ n\pi +{{(-1)}^{n}}\frac{\pi }{2} $
C) $ 2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12} $
D) None
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ \sqrt{3}+1=r\cos \alpha , $ and $ \sqrt{3}-1=rsin\alpha $
$ \therefore r^{2}={{( \sqrt{3}+1 )}^{2}}+{{( \sqrt{3}-1 )}^{2}}=8 $ i.e. $ \alpha =\pi /12 $ From the equation, $ r\cos (\theta -\alpha )=2 $
$ \Rightarrow \cos (\theta -\pi /12)=1/\sqrt{2}=\cos (\pi /4) $
$ \therefore ,\theta =2n\pi \pm \pi /4+\pi /12 $
BETA
