Let $p$ and $q$ be real numbers such that:
$$( \tan p - 1)^3 + 2026(\tan p - 1) = -1 \quad\text{and}\quad (1 - \cot q)^3 + 2026(1-\cot q) = -1$$
with $\tan p \neq \cot q$. Find the number of possible values of $r$ satisfying $\tan p + \cot q + \sin r + \cos r = 3$ in $x\in[-2\pi, 2\pi]$.
Step-by-Step Solution
Key Concept: Let $f(t)=t^3+2026t$. Both equations say $f(\tan p-1)=-1$ and $f(1-\cot q)=-1$. Since $f$ is strictly increasing and odd-like, there is a unique solution. Setting $x=\tan p-1$ and $y=1-\cot q$: $f(x)=f(y)=-1\Rightarrow x=y\Rightarrow\tan p-1=1-\cot q\Rightarrow\tan p+\cot q=2$. So $\sin r+\cos r=1$.
$\sin r+\cos r=1$ has solutions $r\in\{-2\pi,\,-3\pi/2,\,0,\,\pi/2,\,2\pi\}$ in $[-2\pi,2\pi]$: $\mathbf{5}$ values.
Correct Answer: 5