5. Solve the equation $[\sin x]^{2}=\cos ^{2} x-1$, where $[a]$ denotes the integer part of the number $a$.

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Solution. The integer $[\sin x]$ can only take the values $0, 1, -1$.

1) If $[\sin x]=0$, then $\cos ^{2} x=0+1=1$, i.e., $\cos x= \pm 1$. Solving this equation, we find $x=\pi n$. At these points, $\sin ^{2} x=0$. The equality is true.
2) If $[\sin x]= \pm 1$, then $\cos ^{2} x=( \pm 1)^{2}+1=2$, which is impossible. The answer will only be the first case.

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