Solve the equation $2 \sin \pi x / 2 - 2 \cos \pi x = x^{5} + 10 x - 54$.

#
Transferring all terms to one side, we get the equation $x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x=0$. Consider the function

$f(x)=x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x$. Note that $f^{\prime}(x)=5 x^{4}+10-\pi \cos \pi x / 2-2 \sin \pi x>5 x^{4}+10-3 \pi>0$. Therefore, $f(x)$ is an increasing function, so the equation $f(x)=0$ has no more than one root.

Since $f(2)=32+20-54-2 \sin \pi+2 \cos 2 \pi=0$, then $2$ is a root of the original equation.