#### Answer

The function $f(x)$ is continuous on $[0, \dfrac{\pi}{2} ]$ with $f(0)=-1$ and $f(\pi)=(\dfrac{\pi}{2})^k \gt 0$
Therefore, by the Intermediate Value Theorem, there is a $c \in[0, \dfrac{\pi}{2} ]$ such that $f(c)= c^k-\cos (c)=0 \implies \cos c=c^k$. So, the equation $cos k =x^k$ has a solution $c$ in $[0, \dfrac{\pi}{2} ]$.

#### Work Step by Step

We are given that $f(x)=x^k-\cos k$
The function $f(x)$ is continuous on $[0, \dfrac{\pi}{2} ]$ with $f(0)=-1$ and $f(\pi)=(\dfrac{\pi}{2})^k \gt 0$
Therefore, by the Intermediate Value Theorem, there is a $c \in[0, \dfrac{\pi}{2} ]$ such that $f(c)= c^k-\cos (c)=0 \implies \cos c=c^k$. So, the equation $cos k =x^k$ has a solution $c$ in $[0, \dfrac{\pi}{2} ]$.