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 1

The correct option is B no real roots.

Let $e^{\sin x}=t$
Then equation would be
$t-\frac{1}{t}-4=0\cdots \left(1\right)$
$t-\frac{1}{t}$ is an increasing function because its derivative is always positive except for zero where it is not defined. Now for equation $\left(1\right)$ to have real roots, $t-\frac{1}{t}$ must be equal to $4$.
Since $t$ is periodic, the domain of $t-\frac{1}{t}$ will be $[\frac{1}{e},e]$.
Hence, its range will be $[\frac{1}{e}-e,e-\frac{1}{e}]$.
Clearly, the range of $t-\frac{1}{t}$ does not contain $4$.
Hence, equation $\left(1\right)$ will have no real roots.