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

The correct option is B no real roots

Given equation is$e^{sinx}-e^{sinx}-4=0$
Put $e^{sinx}=t$ in the given equation, we get
$t^2-4t-1=0$
$\Rightarrow t=\frac{4\pm \sqrt{16+4}}{2}=\frac{4\pm \sqrt{20}}{2}=\frac{4\pm 2\sqrt{5}}{2}=2\pm \sqrt{5}$
$\Rightarrow e^{sinx}=2\pm \sqrt{5}(\therefore t=e^{sinx})$
$\Rightarrow e^{sinx}=2-\sqrt{5}$ and $e^{sinx}=2+\sqrt{5}$
$\Rightarrow e^{sinx}=2-\sqrt{5}<0$ and $sinx=ln(2+\sqrt{5})>1$
So rejected,
Hence given equation has no solution.
$\therefore$ The equation has no real roots.