15. The equation $(\arccos x)^{2}+(2-t) \arccos x+4$ $=0$ has real solutions. Then the range of values for $t$ is $\qquad$.
15. $[6,+\infty)$.

Let $a=\arccos x$. Then the original equation can be transformed into $a^{2}+(2-t) a+4=0$, which has real solutions in $[0, \pi]$.
By $\Delta=t^{2}-4 t-12 \geqslant 0$, solving yields $t \geqslant 6$ or $t \leqslant-2$.
Since the product of the two roots is 4, it is known that the two roots have the same sign, so the sum of the two roots $t-2>0$, solving yields $t>2$. Therefore, $t \geqslant 6$. At this point, at least one of the roots of the equation is in $[0, \pi]$, otherwise both roots are greater than $\pi$, then the product of the two roots would be greater than 4. Contradiction.