Answer
The smallest non-negative solution is $x = \frac{\pi}{2}$
In general, the non-negative solutions have the form: $x = \frac{\pi}{2} + 2\pi~n$, where $n$ is a non-negative integer
Work Step by Step
$csc~x-cot~x = 1$
$\frac{1}{sin~x}-\frac{cos~x}{sin~x} = 1$
$1-cos~x = sin~x$
$1-2~cos~x+cos^2~x = sin^2~x$
$sin^2~x+cos^2~x-2~cos~x+cos^2~x = sin^2~x$
$2~cos^2~x-2~cos~x = 0$
$2~cos^2~x = 2~cos~x$
$cos^2~x = cos~x$
Then $cos~x = 0$ or $cos~x = 1$
If $cos~x = 1$, then $x = 0$. However, then $csc~x$ is undefined in the original equation. Thus $x = 0$ is not a valid solution.
Therefore, $cos~x = 0$.
Then in the original equation:
$csc~x-cot~x = 1$
$csc~x-0 = 1$
$csc~x = 1$
$sin~x = 1$
$x = arcsin(1)$
$x = \frac{\pi}{2}$
The smallest non-negative solution is $x = \frac{\pi}{2}$
In general, the non-negative solutions have the form: $x = \frac{\pi}{2} + 2\pi~n$, where $n$ is a non-negative integer
