Given, $x(n)=8+2 \sin \left(\omega_{0} n\right)+5 \cos \left(\omega_{0} n\right)+\cos \left(2 \omega_{0} n+\frac{\pi}{2}\right)$

By, Euler's formula, x(n) can be written as-

$x(n)=8+2\left(\frac{e^{j \omega_0 n}-e^{-j \omega_0 n}}{2 j}\right)+5$ $\left(\frac{e^{j \omega_0 n}+e^{-j \omega_0 n}}{2}\right)+$ $\left(\frac{e^{j\left(2 \omega_0 n+\frac{\pi}{2}\right)}+e^{-j\left(2 \omega_0 n+\frac{\pi}{2}\right)}}{2}\right)$

$x(n)=8+\left(\frac{5}{2}+\frac{1}{j}\right) e^{j \omega_0 n}+$ $\left(\frac{5}{2}-\frac{1}{j}\right) e^{-j \omega_0 n}+\left(\frac{1}{2} e^{j \frac{\pi}{2}}\right)$ $e^{j 2 \omega_0 n}+\left(\frac{1}{2} e^{-j \frac{\pi}{2}}\right) e^{-j 2 \omega_0 n}$

Comparing the equation with the equation

$x(n)=C_0+C_1 e^{j \omega_0 n}$ $+C_{-1} e^{-j \omega_0 n}+$ $C_2 e^{j 2 \omega_0 n}+C_{-2} e^{-j 2 \omega_0 n}$

Then,

C₀ = 8

$\begin{aligned} &C_{1}=\left(\frac{5}{2}+\frac{1}{j}\right) \\ &C_{-1}=\left(\frac{5}{2}-\frac{1}{j}\right) \\ &C_{2}=\frac{1}{2} e^{j \pi / 2}=\frac{1}{2} j \\ &C_{-2}=\frac{1}{2} e^{-j \pi / 2}=-\frac{1}{2} j \end{aligned}$

$\begin{aligned} &\left|C_{1}\right|=\left|C_{-1}\right|=\sqrt{\left(\frac{5}{2}\right)^{2}+(1)^{2}} \\ &=\sqrt{\frac{25}{4}+1} \end{aligned}$

$\left|C_{1}\right|=\left|C_{-1}\right|=\frac{\sqrt{29}}{2}$

Hence, the correct option is (C).

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