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

$e^{sinx}-e^{-sinx}-4=0$

$\Rightarrow e^{2sinx}-4e^{sinx}-1=0$

$\Rightarrow e^{sinx}=2\pm \sqrt{5}$

If $e^{sinx}=2+\sqrt{5}$

$\Rightarrow sinx=ln(2+\sqrt{5})$ ; which is not possible

because $ln(2+\sqrt{5})>1$

If $e^{sinx}=2-\sqrt{5}$

$\Rightarrow sinx=ln(2-\sqrt{5})$; which is not possible because $(2-\sqrt{5})<0$

$\therefore$ No real roots exist