5. Solve the equation $2^{[\sin x]}=3^{1-\cos x}$, where $[a]$ denotes the integer part of the number a.
Solution. The integer $[\sin x]$ can only take the values $0, \pm 1$.

1) If $[\sin x]=0$, then $2^{0}=1$. Therefore, $3^{1-\cos x}=1$, from which $\cos x=1$. The solution to this equation is $x=2 \pi n$. Note that $\sin x=0$ for such $x$.
2) If $[\sin x]=-1$, then $3^{1-\cos x}=2^{-1}=1 / 2$. But $1-\cos x \geqslant 0$, hence $3^{1-\cos x} \geqslant 1$. Therefore, the equation $3^{1-\cos x}=1 / 2$ has no solutions.
3) If $[\sin x]=1$, then $x=\pi / 2+2 \pi k$. For these values, $\cos x=0$. Substituting into the equation gives $2^{1}=3^{1}$, which is impossible.

Answer. $\quad x=2 \pi n, n \in \mathbb{Z}$.