Answer
Discontinuities at $x = \frac{\pi}{6} + 2\pi k$ and $ \frac{5\pi}{6} + 2\pi k$, where $k$ is an integer.
Work Step by Step
Discontinuities exist in $f(x)$, where the denominator of $f(x)$ equals zero. Thus:
$$1-2sin(x) = 0$$
$$sin(x) = \frac{1}{2}$$
We then find that discontinuities exist at:
$x = \frac{\pi}{6} + 2\pi k$ and $ \frac{5\pi}{6} + 2\pi k$
where $k$ is an integer.
