Trigonometric Identities Question 270
Question: If $ 0\le x\le 2\pi , $ then number of roots of equation $ {e^{\sin x}}-{e^{-\sin x}}=4 $ is
Options:
A) 0
B) 1
C) 2
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
The given equation can be written as $ {e^{\sin x}}=4+\frac{1}{{e^{\sin x}}} $ ?.(1) Now $ -1\le \sin x\le 1 $ and $ e<3 $
$ \Rightarrow {e^{\sin x}}<3 $
$ \Rightarrow $ Again as we always have $ \frac{1}{{e^{\sin x}}}>0 $
$ \therefore 4+\frac{1}{{e^{\sin x}}}>4 $ Thus the L.H.S of (1) $ <3 $ and R.H.S of $ (1)>4. $ Hence there is no real values of x which satisfy (1). It follows that the given equation has no real solution.
BETA
