11.2. Find the set of values of the function $f(x)=\sin ^{2} x+\cos x$.
Answer: the set of values of the function is the interval $[-1 ; 5 / 4]$.

On the one hand, $f(x)=\sin ^{2} x+\cos x \geq 0-1=-1$ and this estimate is achieved $f(\pi)=-1$.

On the other hand, $f(x)=1+\cos x-\cos ^{2} x=5 / 4-(1 / 2-\cos x)^{2} \leq 5 / 4$. This estimate is achieved $f(\pi / 3)=5 / 4$.

Thus, the function satisfies the inequality $-1 \leq f(x) \leq 5 / 4$, and takes all intermediate values on the interval $[\pi / 3 ; \pi]$, since it is continuous.

Second solution. Note that the set of values of $f(x)=1+\cos x-\cos ^{2} x$ coincides with the set of values of the function $g(t)=1+t-t^{2},-1 \leq t \leq 1$. The function $g(t)$ is continuous on the interval $[-1 ; 1]$, so its set of values is the interval $\left[\min _{t \in[-1,1]} g(t), \max _{t \in[-1,1]} g(t)\right]$. We have: $g^{\prime}(t)=1-2 t$, the function on $[-1 ; 1]$ has a unique critical point $t=\frac{1}{2}$. Let's find the values of the function at the critical point and at the ends of the interval: $g(-1)=-1 ; g\left(\frac{1}{2}\right)=\frac{5}{4} ; g(1)=1$. It follows that $\min _{t \in[-1,1]} g(t)=-1$ and $\max _{t \in[-1,1]} g(t)=\frac{5}{4}$, and the desired set of values is $\left[-1 ; \frac{5}{4}\right]$. Note that the set of values could also be found by plotting the graph of the quadratic function $g(t)=1+t-t^{2},-1 \leq t \leq 1$.

Comment. Only the answer -1 point.