Question

$e^{\sin x}-e^{-\sin x}=4$ then find the number of real solutions.

Answer 1

We have,

$e^{\sin x}-e^{-\sin x}=4$

Let $e^{\sin x}=y$

$y-y^{-1}=4$

$y-\frac{1}{y}=4$

$y^2-1=4y$

$y^2-4y-1=0$

Compare that and we get,

$a{y}^2+by+c=0$

$a=1,b=-4,c-1$

Using quadratic equation formula,

Then,

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$y=\frac{4\pm \sqrt{16-4\times 1\times (-1)}}{2\times 1}$

$y=\frac{4\pm \sqrt{20}}{2}$

$y=2\pm \sqrt{5}$

Now, $e^{\sin x}=2\pm \sqrt{5}$

$-1\le \sin x\le 1$

$\frac{1}{e}\le e^{\sin x}\le e$

Hence, this is the answer.