We have that the range of $f(x)=\cos^{2n}(x)+\sin^{2n}(x),\; n\in \mathbb{N},\; n\geq 2,\;x\in \mathbb{R}$ is
$$f(x)\in[2^{1-n},1]$$
since with $t=\cos^2(x)$ such that $0\le t\le 1$, then
$$t^n+(1-t)^n.$$
The stationary points are the roots of
$$t^{n-1}-(1-t)^{n-1}=0$$ or $$t=\frac12.$$
What about the range of $g(x)=\cos^{2n+1}(x)+\sin^{2n+1}(x)?$
I have to find $b$ that $(b+1) f(x)-b g(x)=1$
Set $g(x)=\cos^{2n+1}{x}+\sin^{2n+1}{x}$ Then $$g'(x)=0\iff \cos^{2n}{x}\sin{x}=\sin^{2n}x\cos{x}\tag{1}$$ Clearly, the range of $g$ contains the interval $[-1,1]$ and we seek values outside this interval. In analyzing $(1)$ therefore, we may assume $\sin{x}\neq0,\, \cos{x}\neq0.$ Then $(1)$ becomes $$\sin^{2n-1}{x}=\cos^{2n-1}{x}$$ so that we must consider the points $x$ such that $$\sin{x}=\cos{x}=\pm{1\over\sqrt{2}}$$ For such $x$, $$|g(x)|=2\cdot\left(2^{-1/2}\right)^{2n+1}=2^{(1-2n)/2}$$ If $n>0$ this value is less than $1$, but when $n=0$ it is greater than $1$. $$\operatorname{Rng}(g)=\cases{\left[-\sqrt{2},\sqrt{2}\right],&$n=0$\\ [-1,1],&$n>0$}$$
