I got as far as this
$\cot(\frac{\pi}{2} – \pi \cdot \cot(x)) = \cot( \pi \cdot \tan(x))$
But when I simplify it further I eventually get $\cot(x) + \tan(x) = \frac{1}{2}$
Which has no solutions.
I have a feeling I need to do something with this:
$\tan(mx) = \cot(nx)$ and general solution to this is $x = \frac{1}{m+n}(r\pi + \pi/2)$
The solution is probably really obvious as usual
TIA
Best Answer
After your simplification, your equation is equivalent to $$\pi\tan x=\frac\pi2-\pi\cot x+k\pi\iff \tan x+\cot x=\frac12+k\iff \frac{2}{\sin2x}=\frac12+k$$ So the solutions are the valid (and non null) solutions of $$\sin 2x=\frac{4}{1+2k}$$ $k\in\mathbb Z$ has to verify $1+2k\ge4$ or $1+2k\le -4$, so $k\ge2$ or $k\le -3$. So the solutions are $$S=\left\{\frac{1}{2}\arcsin\frac{4}{1+2k}+l\pi,\ k\le -3\text{ or }k\ge2,\ l\in\mathbb Z\right\}\cup \left\{l\pi-\frac{1}{2}\arcsin\frac{4}{1+2k},\ k\le -3\text{ or }k\ge2,\ l\in\mathbb Z\right\}$$