If the difference between the greatest and the least value of the function $f(x)={\sin }^{2}x+\cos x$ on $[0,\pi ]$ is $\frac{a-4}{4}$, then $a$ equals:

$f(x)={\sin }^{2}x+\cos x=(1-{\cos }^{2}x)+\cos x=-{\cos }^{2}x+\cos x+1$.Let $t=\cos x$. Since $x\in [0,\pi ]$, $t\in [-1,1]$.The function becomes $g(t)=-t^2+t+1$.Differentiate $g(t)$: $g^{\prime }(t)=-2t+1$.Stationary point: $-2t+1=0⟹t=1/2$.Evaluate $g(t)$ at boundaries and stationary point:$g(1)=-(1{)}^2+1+1=1$ (occurs at $x=0$)$g(-1)=-(-1{)}^2+(-1)+1=-1$ (occurs at $x=\pi$)$g(1/2)=-(1/2{)}^2+1/2+1=-1/4+1/2+1=5/4$ (occurs at $x=\pi /3$)Greatest value $M=5/4$, Least value $m=-1$.Difference $M-m=5/4-(-1)=9/4$.Given $\frac{a-4}{4}=\frac{9}{4}⟹a-4=9⟹a=13$.