Solve $\sec(x)=\tan(x)$

Question

I have been trying to do this problem:

Solve $$\sec(x)=\tan(x),\quad 0≤x<2π$$

I started by rewriting $\sec(x)$ as $\frac{1}{\cos(x)}$.

I then rewrote $\tan(x)$ to get $\frac{\sin(x)}{\cos(x)}$.

Therefore:

$$\frac{1}{\cos(x)} = \frac{\sin(x)}{\cos(x)}$$
Therefore:

$$1=\sin(x)$$

$$x=\frac{π}{2}$$

However, this is not a solution to the original problem because both $y=\tan(x)$ and $y=\sec(x)$ have asymptotes as $x=\frac{π}{2}$

So where have I gone wrong, or are there no solutions?

If there are no solutions, how would I know (in other words, can we prove there are no solutions, or is $x=\frac{π}{2}$ not working enough to conclude there are no solutions?)

Answer 1

$$\sec^2x-\tan^2x=1$$ It fails at the solution of this problem. Hence there is no solution, because this identity never fails.

Hope it helps:)