Question

The equation $e^{sinx} - e^{-sinx} - 4 = 0$ has

• A. infinite number of real roots
• B. no real roots
• C. exactly one real root
• D. exactly four real roots

Answer 1

Answer: (B)

Let $e^{sinx} = t$
$\Rightarrow \quad t^{2} - 4t - 1 = 0$
$\Rightarrow \quad t = \frac{4\pm\sqrt{16+4}}{2}$
$\Rightarrow \quad t = e^{sinx} = 2 \pm \sqrt{5}$
$\Rightarrow \quad e^{sin \,x} = 2 -\sqrt{5} ,\quad\quad e^{sin \,x} = 2 + \sqrt{5}$
$e^{sin \,x} = 2 - \sqrt{5}< 0,\quad\quad\Rightarrow \quad sinx = ln\left(2 +\sqrt{5}\right) > 1$
so rejected $\quad\quad$ so rejected
hence no solution