If we have a function $f$ then at the maxima and minima of $f$, we have $f'=0$.
$$f(\theta)=\cos(x)\implies f'(\theta)=-\sin(\theta)$$
We know that:
$$-\sin(\theta)=0\implies \sin(\theta)=0\implies \theta=\pi k,k\in\Bbb{Z}$$
Since we have a domain $-\pi\leq\theta\leq\dfrac{\pi}{6}$, we have maxima or minima at $0\pi=0$ and $-1\pi=-\pi$. And:
$$\cos(0)=1$$
$$\cos(-\pi)=-1$$
So we have a maximum at $(0,1)$ and a minimum at $(-\pi,-1)$. Since there are no other multiples of $\pi$ in the interval $-\pi\leq\theta\leq\dfrac{\pi}{6}$ and therefore no other solutions to $-\sin(\theta)=\theta$ in the interval, we are done.
From your calculation, you have shown that there is no critical point in the interior of the half-disk, but the left endpoint of the semicircular boundary is one. The index of that point is
$$ D \ \ = \ \ f_{xx}·f_{yy} \ - \ (f_{xy})^2 \ \ = \ \ 2·(-2) \ - \ 0^2 \ = \ -4 \ \ < \ \ 0 \ \ , $$
which identifies $ \ (-1 \ , \ 0 ) \ $ as a saddle-point. This makes sense since the function is "concave upward" in the $ \ x-$ direction about $ \ x \ = \ -1 \ $ , due to the $ \ x^2 \ + \ 2x \ = \ (x + 1)^2 \ - \ 1 \ $ terms, but is "concave downward" in the $ \ y-$ direction because of the $ \ -y^2 \ $ term. As we will need it for comparisons later, we compute that $ f(-1,0) \ = \ (-1)^2 \ + \ 2(-1) \ - \ 0^2 \ = \ -1 \ \ . $
The lower boundary of the region is the $ \ x-$ axis over the interval $ \ [ \ -1 \ , \ 1 \ ] \ $ . Inserting $ \ y \ = \ 0 \ $ into the function expression gives $ \ f(x,0) \ = \ (x + 1)^2 \ - \ 1 \ \ . $ We have already found $ \ f(-1,0) \ $ to be the minimum on this line segment; the right endpoint has the value $ f(1,0) \ = \ (1+1)^2 \ - \ 1 \ = \ 3 \ \ . $
It remains to examine the upper semicircular boundary. There are a number of ways we might set this up, but to avoid having to work with a square-root or trigonometric identities, we will use the circle equation to write $ \ y^2 \ = \ 1 \ - \ x^2 \ $ and express our function on this section of the boundary as
$$ \ \phi(x) \ = \ (x + 1)^2 \ - \ 1 \ - \ (1 - x^2) \ \ = \ \ (x + 1)^2 \ + \ x^2 \ - \ 2 \ \ . $$
[This gives us the correct values for the function at the endpoints of the region:
$$ \phi(-1) \ \ = \ \ ([-1] + 1)^2 \ + \ (-1)^2 \ - \ 2 \ \ = \ -1 \ \ \ , \ \ \ \phi(1) \ \ = \ \ (1 + 1)^2 \ + \ 1^2 \ - \ 2 \ \ = \ \ 3 \ \ . \ ] $$
We find $ \ \phi'(x) \ = \ 2·(x + 1) \ + \ 2x \ \ = \ \ 0 \ \ \ \Rightarrow \ \ \ 4x \ \ = \ -2 \ \ \ \Rightarrow \ \ \ x \ = \ -\frac{1}{2} \ \ , $ for which $ \phi \left( -\frac{1}{2} \right) \ \ = \ \ \left( \ \left[-\frac{1}{2} \right] + 1 \ \right)^2 \ + \ \left(-\frac{1}{2} \right)^2 \ - \ 2 \ \ = \ \ \frac{1}{4} \ + \ \frac{1}{4} \ - \ 2 \ \ = \ -\frac{3}{2} \ \ . $
Hence, we observe that the absolute maximum of the function on this closed half-disc is $ f(1,0) \ = \ 3 \ $ and the absolute minimum is $ f \left( -\frac{1}{2} \ , \ \frac{\sqrt{3}}{2} \right) \ = \ -\frac{3}{2} \ \ . $ The saddle point at $ \ (-1 \ , \ 0 ) \ $ has no special role in this region.
Best Answer
Since $$ \nabla f(x,y)=(0.0) \iff (x,y) =(\pm2.0) \notin D, $$ the maximum and minimum of $f$ on $D$ can only be achieved on the boundary $\partial D$.
The boundary consists of $$ L=\{(x,y): x=1, |y| \le 1\} $$ and $$ C=\{(x,y): (x-1)^2+y^2=1, 0\le x\le 1\}=\varphi\left(\left[\frac{3\pi}{2}m\frac{\pi}{2}\right]\right), $$ with $$ \varphi(t)=(1+\cos t, \sin t). $$ THe restrection, $f_{|L}$ of $f$ on $L$ given by $$ f(1,y)=y^2-22, $$ satisfies $$ f(1,0)=-22=f(1,0)\le f_{|L}*x,y)\le -21=f(1,\pm1) \quad \forall (x,y)\in L. $$
To find the maximum/minimum of $f$ ob $C$, we can use the method of Lagrange multipliers.
Consider the function defined by
$$ F(x,y,\lambda)=f(x,y)-\lambda [(x-1)^2+y^2-1]. $$ At a critical point $(x,y) \in C$, the gradient of $F$ [w.r.t $(x,y)$] must be $(0,0)$, i.e.
\begin{eqnarray} 6x^2-24 &=& 2\lambda(x-1)\cr 2y&=&2\lambda y \end{eqnarray} We have $$ 2y=2\lambda y \iff \lambda=1 \quad or \quad y=0. $$
If $\lambda =1$, thrn $$ 6x^2-24=2(x-1) \iff x=\frac{-1\pm\sqrt{133}}{6} \notin [0,1] $$
If $y=0$, then thanks to the constraint we get $x=0$ or $x=1$. In particular, if $x=0$, then $\lambda=12$.
Since $f(0.0)=0$, we have $$ \min_{D}f=-22,\quad \max_{D}f=0. $$