Trigo Range

It's tempting to eliminate $x$ or $y$ using the condition, but this leads to a lot of case analysis. Instead, we can use Lagrangian multipliers: define $f(x,y,L)=x-\cos y + L \left(x^2+2\cos y-1 \right)$

By setting the partial derivatives to zero, we can find the extreme values subject to the constraint. Differentiating: $\begin{aligned} f_x&=1+2Lx \\ f_y&=(1-2L)\sin y \\ f_L&=x^2+2\cos y-1 \end{aligned}$

From $f_y=0$ , either $\sin y=0$ or $L=\frac12$ .

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Case $\sin y=0$ :

In this case, $\cos y=\pm1$ . From $f_L=0$ , $x^2\pm2-1=0$

For this to have a real solution, we need $\cos y=-1$ ; hence $x^2-3=0$

and $x=\pm \sqrt3$ .

When $x=\sqrt3$ , $x-\cos y=1+\sqrt3$ . When $x=-\sqrt3$ , $x-\cos y=1-\sqrt3$

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Case $L=\frac12$ :

From $f_x=0$ , we find $x=-1$ .

Now from $f_L=0$ , we get $\cos y=0$ ; so $x-\cos y=-1$ .

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The largest of these values is $1+\sqrt3$ ; the smallest is $-1$ ; so the range is $\boxed{\left[-1,1+\sqrt3 \right]}$ .

It would be helpful if the expression $x - \cos(y)$ was in terms of $x$ . By solving for $\cos(y)$ in the given equation we find that $\displaystyle \cos(y) = \frac{1 - x^2}{2}$ . Since $\cos(y)$ has a range of $[-1, 1]$ we must conclude that the domain of $x$ is $[-\sqrt3, \sqrt3]$ .

Substituting our expression for $\cos(y)$ in terms of $x$ into $x - \cos(y)$ yields $\displaystyle \frac{x^{2}+2x-1}{2}$ . This is a parabola with a minimum at $(-1, -1)$ and a maximum at $(\sqrt3, 1 + \sqrt3)$ because the domain of $x$ is bounded. The range of $x - \cos(y)$ is therefore $\boxed{[-1, \sqrt3 + 1]}$ .

From $x^2 + 2\cos y = 1$ , $\implies x = \pm \sqrt{1-2\cos y}$ . Then $x - \cos y = \pm \sqrt{1-2\cos y} - \cos y$ . Note that $\sqrt{1-2\cos y}$ is real for $\frac \pi 3 \le y \le \frac {5\pi}3$ .

Consider the maximum value of $x - \cos y = \pm \sqrt{1-2\cos y} - cos y$ . Since both $\sqrt{1-2\cos y}$ and $(-\cos y)$ are maximum when $y=\pi$ , then $\max(x-\cos y) = \sqrt{1-2\cos \pi} - \cos \pi = \sqrt 3 + 1$ .

Now consider the minimum value. Since $\sqrt{1-2\cos y} \ge 0$ , $\min(x-\cos y)$ occurs when $x-\cos y = - \sqrt{1-2\cos y} - \cos y$ or when $\sqrt{1-2\cos y} + \cos y$ is maximum. Let $z = \sqrt{1-2\cos y} + \cos y$ . To find the maximum,

$\begin{aligned} \dfrac {dz}{dy} & = \dfrac {\sin y}{\sqrt{1-2\cos y}} - \sin y & \small \blue{\text{Putting }\frac {dz}{dy}=0} \end{aligned}$

$\sin y \left(\dfrac 1{\sqrt{1-2\cos y}}-1\right) = 0 \implies \begin{cases} \sin y = 0 & \implies y = \pi & \implies z = \sqrt 3 - 1 \\ \dfrac 1{\sqrt{1-2\cos y}}-1 = 0 & \implies y = \frac \pi 2 & \implies z = 1\end{cases}$

Therefore $\min(x-\cos y) = - 1$ and the range of $x - 2\cos y$ is $\boxed{\left[-1, \sqrt 3+1\right]}$ .