Question
The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has :
A.
infinite number of real roots
B.
no real roots
C.
exactly one real root
D.
exactly four real roots
Answer :
no real roots
Solution :
Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$
put $${e^{\sin x}} = t$$ in the given equation, we get
$$\eqalign{
& {t^2} - 4t - 1 = 0 \cr
& \Rightarrow \,\,t = \frac{{4 \pm \sqrt {16 + 4} }}{2} \cr
& = \frac{{4 \pm \sqrt {20} }}{2} \cr
& = \frac{{4 \pm 2\sqrt 5 }}{2} \cr
& = 2 \pm \sqrt 5 \cr
& \Rightarrow \,\,{e^{\sin x}} = 2 \pm \sqrt 5 \left( {\because \,t = {e^{\sin x}}} \right) \cr
& \Rightarrow \,\,{e^{\sin x}} = 2 - \sqrt 5 \,\,{\text{and }}{e^{\sin x}} = 2 + \sqrt 5 \cr
& \Rightarrow \,\,{e^{\sin x}} = 2 - \sqrt 5 \, < 0\,{\text{and }}\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1 \cr
& \,{\text{So rejected }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{So rejected}} \cr} $$
Hence given equation has no solution.
∴ The equation has no real roots.