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?)
$\endgroup$2 Answers
$\begingroup$There is nothing wrong with what you did. The problem has no solution since, precisely, if $x\in\mathbb R$ is such that both $\tan(x)$ and $\sec(x)$ are defined, then $\sec(x)=\tan(x)\iff\sin(x)=1$. But at those numbers $x$ for which $\sin(x)=1$, $\tan(x)$ and $\sec(x)$ are undefined.
$\endgroup$ 3 $\begingroup$$$\sec^2x-\tan^2x=1$$It fails at the solution of this problem. Hence there is no solution, because this identity never fails.
Hope it helps:)
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