Since $-1 \leq x \leq 1$
$\therefore \quad-\frac{\pi}{4} \leq \tan ^{-1} x \leq \frac{\pi}{4} \quad \therefore \quad\left[\tan ^{-1} x\right]=-1,0$
When $\left[\tan ^{-1} x\right]=-1$, then $\left\{\cos ^{-1} x\right\}=1 \quad$ (not possible)
When $\left[\tan ^{-1} x\right]=0$, then $\left\{\cos ^{-1} x\right\}=0 \quad \therefore \quad \cos ^{-1} x$ is integer
Since $0 \leq \cos ^{-1} x \leq \pi \quad \therefore \quad \cos ^{-1} x=0,1,2,3$
$x=\cos 0, \cos 1, \cos 2, \cos 3$ but $x * \cos 2, \cos 3 \quad \therefore \quad$ the solution set is $\{1, \cos 1\}$
