$x(n) = \sin(0.25\pi n + 0.4)$

Solution:

Consider Standard equation $x(n) = A \cos(\pi n+\theta)$

Comparing given signal we get, $\omega n = 0.25 \pi n \\ \therefore 2\pi fn = \dfrac {\pi n}{4} \\ \therefore f = \dfrac18$

**Result : Given signal, x(n) is periodic with Fundamental Period,N = 4**

$x(n) = cos(0.5n\pi) + sin(0.25n\pi)$

Solution:

Consider x(n) to be combination of $x_1(n)$ & $x_2(n)$

Where, $x_1(n)= \cos(0.5 \pi n_1) \& \ x_2(n)=\sin(0.25 \pi n_2)$

$\therefore, \omega_1n_1 = 0.5\pi n1 \ \ \ \ \& \omega_2n_2 = 0.25 \pi n2$

$\therefore, 2 \omega f_1n_1 = \dfrac{1}{2} \pi n1 \ \ \ \& 2 \pi f_2n_2 = \dfrac{1}{4} \pi n_2$

$\therefore, f_1 = \dfrac{1}{4} \ \ \ \& f_2 = \dfrac{1}{8}$

$\therefore, N_1 = 4 \ \ \ \ \& N_2 = 8$

$\therefore, N = LCM of N_1 \ \ \& N_2 = 8$

**Result : Given signal, $X_2 (n)$ is periodic with Fundamental Period, N = 8**