Solving equation $\tan(5π\cos\alpha) = \cot(5π\sin\alpha)$ $$\tan(5π\cos \alpha) = \cot(5π\sin 
\alpha)$$
I did that $\tan(5π\cos\alpha) = \tan\left[\frac π2-5π\sin\alpha\right]$
And then used the solution of Trigonometric Equation $\tan(\theta)=\tan(\beta)$
Which is
$\theta = nπ + \beta$, $n$ is an integer.
But the basic condition of using the above result is that $\beta$ lies between $\left(-\frac π2,\frac π2\right)$
And so gives $\sin \alpha $ lies between $\left(0,\frac 15\right)$
What is wrong with this?
PS correct answer comes by using my method..
 A: Using the formula in the question, we get $$5\pi\cos\alpha=n\pi+\frac \pi2-\sin\alpha$$Where n is an integer.
Simplifying, we get $$\sin\alpha+\cos\alpha=\frac{2n+1}{10}$$
Now, there are many ways to show that $\sin\alpha+\cos\alpha=\sqrt2\sin(\alpha+\frac\pi4)$. I'm not going to prove that here. So, we have$$\sin(\alpha+\frac\pi4)=\frac{2n+1}{10\sqrt2}$$Now, moving the sine to the other side and subtracting $\frac\pi4$ on both sides, we get$$\alpha=\arcsin(\frac{2n+1}{10\sqrt2})-\frac\pi4$$However, this only holds when the argument of the arcsine lies between 1 and -1. Or,$$-1\leq\frac{2n+1}{10\sqrt2}\leq1$$solving this, we get$$\frac{-10\sqrt2-1}{2}\leq n\leq \frac{10\sqrt2-1}{2}$$Combining this with the original restraint that n is an integer, we get $n=0, \pm1, \pm2, \pm3, \pm4, \pm5, \pm6,-7$. Therefore, our final answer is$$\alpha=\arcsin(\frac{2n+1}{10\sqrt2})-\frac\pi4,n=0, \pm1, \pm2, \pm3, \pm4, \pm5, \pm6,-7 $$Its my first time writing an answer here, so I omitted a few simple steps. Hope you don't mind.
A: If $\tan A=\cot B \implies A=n\pi+\pi/2-B \implies A+B=(n+1/2)\pi, n\in I^.$
So here, we have $$5 \pi [\sin \alpha+\cos \alpha] =(n+1/2)\pi\implies \sin [\alpha+\pi/4]=-1 \ge \frac{n+1/2}{5\sqrt{2}} \le 1, n=$$
$$ \implies \alpha= \sin^{-1}\frac{(n+1/2)}{5\sqrt{2}}, n=0\pm 1,\pm 2,\pm 3,\pm 4, \pm 5, \pm 6, -7$$
A: We get $\sin( 5 \pi \cos a)\sin(5\pi \sin a)-\cos (5\pi \cos a) \cos (5\pi \sin a)=0$, which gives $\cos (5\pi \cos a+5\pi \sin a)=0$    or $5\pi \cos a+5\pi \sin a=k\pi+\pi/2$, or $\cos a+\sin a=k/5+1/10$, k is an integer.
