Find $A(x)=a+b\cos(cx+d)$

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Consider the function $A(x)$ of the form $a+b\cos(cx+d)$.We want to determine the coefficients so that the function have the following properties:

1.$A:[0,1]\rightarrow [0,1]$ be a continuous increasing function such that $A(0)=0\ and \ A(1)=1$.

2.There exist a continuous function $U:\mathbb R\rightarrow [0,1]$ with compactly support subject to

$U(x) = \begin{cases} 0 & \text{if $x\lt x_0$ } \\ A(\frac {x-x_0}{x_1-x_0}), & \text{if $x_0\le x \lt x_1$ }\\ 1& \text{if $x=x_1$}\\ A(\frac {x_2-x}{x_2-x_1}), & \text{if $x_1\le x \lt x_2$ }\\ 0 & \text{if $x\ge x_2$ } \end{cases} $.

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Possibly you can define $A(x)$ with
$$\begin{align} a &= \frac {-\cos p}{\cos r - \cos p} \\ b &= \frac 1{\cos r - \cos p} \\ c &= r-p \\ d &= p \end{align}$$ for any $p,\ r$ such that $\pi\le p\lt r\le 2\pi$.

The simplest case is, of course, $$A(x)=\frac{1+\cos(\pi(x+1))}2$$ with $a = b = 1/2,\ c = d = \pi$, obtained for $p=\pi,\ r=2\pi$.