If the number of principal solutions of the equation tan (7π cos x)...

$\tan \left(7\pi cos x\right) = \tan \left(\dfrac{\pi}{2} - 7\pi sin x\right)$
$⇒ 7\pi cos x = n\pi + \dfrac{\pi}{2} - 7\pi sin x$
$⇒ 7\pi ( sin x + cos x ) = \left(n + \dfrac{1}{2}\right)\pi$
$⇒ sin x + cos x = \dfrac{2n + 1}{14}$
$⇒ \sin \left(x + \dfrac{\pi}{4}\right) =\dfrac{2n + 1}{14\sqrt{2}}$
$- 1 \le \sin \left(x + \dfrac{\pi}{4}\right) \le 1$
$- 1 \le \dfrac{2n + 1}{14\sqrt{2}} \le 1$
$\dfrac{- 14\sqrt{2} - 1}{2} \le n \le\dfrac{14\sqrt{2} - 1}{2}$
$- 10.4 \le n \le 9.4$
$⇒ n = - 10, - 9, \ldots \ldots , - 1, 0, 1, \ldots . , 9$
Total $20$ values of $n$ and for each value of $n$ their will be two solutions.

Hence, $k = 40 ⇒ \dfrac{k}{5} = 8$