9. Let the real number $t \in[0, \pi]$, if the equation $\cos (x+t)=1-\cos x$ has a solution for $x$, find the range of values for $t$.

Let the real number $t \in[0, \pi]$, if the equation $\cos (x+t)=1-\cos x$ has a solution for $x$, find the range of values for $t$.
Solution: The original equation is equivalent to $\cos \left(x+\frac{t}{2}\right) \cos \frac{t}{2}=\frac{1}{2}$, $\qquad$
When $t=\pi$, the left side of the equation equals 0, which clearly has no solution;
When $t \in[0, \pi)$, the equation further simplifies to $\cos \left(x+\frac{t}{2}\right)=\frac{1}{2 \cos \frac{t}{2}}$, noting that $\cos \left(x+\frac{t}{2}\right) \in[-1,1]$, and $\cos \frac{t}{2}>0$

Therefore, the equation has a solution if and only if $0<\frac{1}{2 \cos \frac{t}{2}} \leq 1$ $\qquad$
which means $\frac{1}{2} \leq \cos \frac{t}{2} \leq 1$, solving this yields $0 \leq t \leq \frac{2 \pi}{3}$.
In summary, $t \in\left[0, \frac{2 \pi}{3}\right] \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ points