Let $x = \sin(\theta), y = \cos(\theta)$
$$3 - 2 y - 4x - 2y^2+1 + 2xy = 0$$
Simplify, divide by $2$ and replace $y^2$ with $1-x^2$.
$$1 - y - 2x+x^2+ xy = 0$$
Factor
$$(x-1)(x+y-1) = 0$$
Now just solve $\sin(\theta) = 1$ and $\sin(\theta) + \cos(\theta) = 1$.
Replacing $(x, z)$ by $(-x, -z)$ or $(y, z)$ by $(-y, -z)$ does not change the inequality, therefore we can assume that
$$
0 \le x, y \le \pi \, , \, -1 \le z \le 1 \, .
$$
Case 1: $x+y \le \pi$. The solution for this case is inspired by Proving that $\cosh^{-1} (\cosh(x)\cosh(y) ) \geq \sqrt{x^2 + y^2}$ and the CAT(0) inequality.
The function $f(u) = \cos(\sqrt u)$ is convex on the interval $0 \le u \le \pi^2$, this is straightforward to verify, e.g. by calculating the second derivative.
We can assume that $y \le x$. The convexity condition with $\lambda = (1+z)/2$ and $1-\lambda = (1-z)/2$ then gives
$$
\begin{align}
\cos x \cos y + z \sin x \sin y &=
\frac{1+z}2 \cos(x-y) + \frac{1-z}2 \cos(x+y) \\
&= \frac{1+z}2 f((x-y)^2) + \frac{1-z}2 f((x+y)^2) \\
&\ge f \left( \frac{1+z}2 (x-y)^2 + \frac{1-z}2 (x+y)^2\right) \\
&= f( x^2+y^2-2xyz) \\
&= \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right) \, .
\end{align}
$$
$f$ is in fact strictly convex, so that equality holds exactly in the following cases:
- $\lambda = 0$ or $1$, that is if $z = \pm 1$.
- $(x-y)^2 = (x+y)^2$, that is if $x=0$ or $y=0$.
Case 2: $\pi < x+y \le 2\pi$. This case can be reduced to the first case, credit goes to ipst. I have only simplified the solution a bit.
Setting $x' = \pi - x$ and $y' = \pi - y$ we can apply the first case to $(x', y', z)$:
$$
\begin{align}
\cos x \cos y + z \sin x \sin y &= \cos x' \cos y' + z \sin x' \sin y' \\
&\ge \cos \left( \sqrt{x'^2 + y'^2 - 2 x' y' z} \right) \\
&\overset{(*)}{\ge} \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right) \, .
\end{align}
$$
The last inequality $(*)$ needs to be proved. For $\alpha, \beta \in [0, 2 \pi]$ we have
$$
\cos \alpha \ge \cos \beta \iff \alpha \le \beta \le 2\pi -\alpha \, .
$$
Therefore we have to show that
$$
\sqrt{x'^2 + y'^2 - 2 x' y' z} \overset{(1)}{\le} \sqrt{x^2 + y^2 - 2 x y z}\overset{(2)}{\le} 2 \pi - \sqrt{x'^2 + y'^2 - 2 x' y' z} \, .
$$
The left part $(1)$ follows from
$$
(x^2 + y^2 - 2 x y z) - (x'^2 + y'^2 - 2 x' y' z) = 2 \pi(x+y-\pi)(1-z) \ge 0
$$
and the right part $(2)$ follows from
$$
\sqrt{x^2+y^2-2xyz} + \sqrt{x'^2+y'^2-2x'y'z}
\le (x+y) + (x'+y') = 2 \pi \, .
$$
Equality holds in the following cases:
- $z = \pm 1$.
- $x'y' = 0$ and $x+y= \pi$, that is if $(x, y) = (\pi, 0)$ or $(x, y) = (0, \pi)$.
Best Answer
For $0 < d \le \pi/2$ we have $$ \sin(d) = \int_0^d \cos(t)\, dt > d \, \frac{1+\cos(d)}{2} $$ because $\cos$ is concave on that interval. Applying this to $d=y-x$ gives $$ \frac{\sin(y-x)}{y-x} > \frac{1+\cos(y-x)}2 \ge \frac{\cos(x) + \cos(y)}2 $$ because $\cos$ is decreasing on $[0, \pi/2]$.