Answer
On the graph of $y = tan(2~sin~x)$ on the interval $[-\pi, \pi]$, we can see that the four asymptotes are approximately at the points $x = \pm 0.9$ and $x = \pm 2.24$
The four asymptotes are:
$x = -\pi + sin^{-1}(\frac{\pi}{4})$
$x = -sin^{-1}(\frac{\pi}{4})$
$x = sin^{-1}(\frac{\pi}{4})$
$x = \pi - sin^{-1}(\frac{\pi}{4})$
Work Step by Step
On the graph of $y = tan(2~sin~x)$ on the interval $[-\pi, \pi]$, we can see that the four asymptotes are approximately at the points $x = \pm 0.9$ and $x = \pm 2.24$
We can find the exact equations of these asymptotes:
$y = tan(2~sin~x)$
$y = \frac{sin(2~sin~x)}{cos(2~sin~x)}$
There will be an asymptote when $cos(2~sin~x) = 0$
$2~sin~x = -\frac{\pi}{2},\frac{\pi}{2}$
$sin~x = -\frac{\pi}{4},\frac{\pi}{4}$
If $sin~x = -\frac{\pi}{4}$:
$x = sin^{-1}(-\frac{\pi}{4}) = -sin^{-1}(\frac{\pi}{4})$
$x = -\pi + sin^{-1}(\frac{\pi}{4})$
If $sin~x = \frac{\pi}{4}$:
$x = sin^{-1}(\frac{\pi}{4})$
$x = \pi - sin^{-1}(\frac{\pi}{4})$
