Trigonometric Equations Question 380
Question: If $ \sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha ), $ then $ x= $
Options:
A) $ n\pi \pm \frac{\pi }{6} $
B) $ n\pi \pm \frac{\pi }{3} $
C) $ n\pi \pm \frac{\pi }{4} $
D) $ n\pi \pm \frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 3\sin \alpha -4{{\sin }^{3}}\alpha =4\sin \alpha ({{\sin }^{2}}x-{{\sin }^{2}}\alpha ) $
$ \therefore $ $ {{\sin }^{2}}x={{( \frac{\sqrt{3}}{2} )}^{2}} $
Þ $ {{\sin }^{2}}x={{\sin }^{2}}\pi /3 $
Þ $ x=n\pi \pm \pi /3 $ .
BETA
