An unusual trigonometry equation!

We are given $\tan(5\pi\cos x)=\cot(5\pi\sin x)$

Rewrite it as $1-\tan(5\pi\cos x)\tan(5\pi\sin x)=0$

Using identity $\tan(A+B)(1-\tan A \tan B)=\tan A + \tan B$ we get,

$\cot(5\pi\cos x + 5\pi\sin x) = 0$

This on solving for the general solutions we will get

$5\pi(\cos x + \sin x) = (2n+1)\frac{\pi}{2}$ where $n \in Z$

Or

$\cos x + \sin x = \frac{2n+1}{10}$

We know that, $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$ which implies

$-\sqrt{2} \leq \frac{2n+1}{10} \leq \sqrt{2}$

$-14.14 \leq 2n+1 \leq 14.14$

$-7.5.. \leq n \leq 6.5..$

Since $n \in Z$ Therefore , there are $14$ integral values of $n$ satisfying the above inequality.

And for each value of $n$ , we will want to solve the equation

$\cos x + \sin x = a$ for $x \in (0,2\pi)$ where $a$ will be obtained by putting different values of $n$

Solving this we get $2$ values of $x$ for the given interval $(0,2\pi)$

Total solutions = $14 \times 2 = \boxed{2}$

tan(5 pi cos x)=tan (pi/2 - 5 pi sin x) this implies 5 pi cos x = n*pi + pi/2 - sin x

on rearranging, the equation becomes, sin(pi/4 + x) = 1/(5 sqrt(2))*(n+1/2)

max +ve value of n for which real x exist is 6 and min -ve value is -7.

For each value ,we get 2 solutions. Hence total no. of solutions = 14*2 = 28

Well this is not a good method but an alternate method by use of graphing calculator. The curve may be graphed in parts so to separate out asymptotes and intersection. In it-83 Plus I had plotted in intervals of 0.3. .........................0 - 0.3, 0.3 - 0.6,. . . $6 - 2*\pi$ .