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