To solve the equation

\[
\tan x + \cot x = \frac{1}{\sin x} - \cot x,
\]

let's start by rewriting the terms in terms of sine and cosine.

1. Recall that \(\tan x = \frac{\sin x}{\cos x}\) and \(\cot x = \frac{\cos x}{\sin x}\).

So, we can rewrite the left side:
\[
\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}.
\]
Finding a common denominator, we get:
\[
\tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}.
\]

2. Now rewrite the right side:
\[
\frac{1}{\sin x} - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x}.
\]

3. Now the equation becomes:
\[
\frac{1}{\sin x \cos x} = \frac{1 - \cos x}{\sin x}.
\]

4. Cross-multiply to eliminate the fractions:
\[
1 \cdot \sin x = (1 - \cos x) \cdot \sin x \cos x.
\]

5. Since \(\sin x \neq 0\), we can divide both sides by \(\sin x\):
\[
1 = (1 - \cos x) \cos x.
\]

6. Distributing gives:
\[
1 = \cos x - \cos^2 x.
\]

Rearranging gives us a quadratic equation:
\[
\cos^2 x - \cos x + 1 = 0.
\]

7. Solving this quadratic equation using the quadratic formula:
\[
\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2}.
\]

8. This results in:
\[
\cos x = \frac{1 \pm i\sqrt{3}}{2}.
\]