The equation $e^{\sin x}-e^{-\sin x}-4=0$ has

The equation $e^{\sin x}-e^{-\sin x}-4=0$ has
  1. infinite number of real roots
  2. no real roots
  3. exactly one real root
  4. exactly four real roots

Solution

$e^{\sin x}-e^{-\sin x}=4 \quad \Rightarrow e^{\sin x}=t$ $t-\frac{1}{t}=4$ $\begin{array}{ll}t^2-4 t-1=0 & \Rightarrow t=\frac{4 \pm \sqrt{16+4}}{2} \\ \Rightarrow t=\frac{4 \pm 2 \sqrt{5}}{2} & \Rightarrow t=2 \pm \sqrt{5}\end{array}$ $e^{\sin x}=2 \pm \sqrt{5} \quad \quad-1 \leq \sin x \leq 1 \quad \quad \frac{1}{e} \leq e^{\sin x} \leq e$ $e^{\sin x}=2+\sqrt{5}$ not possible $e^{\sin x}=2-\sqrt{5}$ not possible $\therefore$ hence no solution

Asked in: JEE Main 2012 (Offline)

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