If $\cos^2 x - a\sin x + b = 0$ has only one solution in $[0, \pi]$. Then:
$a \in (-\infty, -2] \cup (-1, \infty)$
$a \neq b$
$a = b$
$b \in (-\infty, -2] \cup (-1, \infty)$
Step-by-Step Solution
Key Concept: The key idea is to transform the trigonometric equation into a quadratic in terms of $t = \sin x$, considering its domain $[0,1]$ for $x \in [0, \pi]$. Crucially, one must carefully analyze how the number of roots of $t$ in $[0,1]$ maps to the number of solutions for $x$ in $[0, \pi]$: only $t=1$ corresponds to a single solution for $x$, while $t \in (0,1)$ and $t=0$ both correspond to two solutions for $x$.
Step 1: Rewrite the equation using $\cos^2 x = 1 - \sin^2 x$: $1 - \sin^2 x - a\sin x + b = 0$, i.e., $\sin^2 x + a\sin x - (1+b) = 0$.
Step 2: Let $t = \sin x$. For $x \in [0, \pi]$, $\sin x \geq 0$, so $t \in [0, 1]$. The equation becomes $t^2 + at - (1+b) = 0$.
Step 3: For the original equation to have only one solution in $[0, \pi]$, the quadratic $f(t) = t^2 + at - (1+b)$ must have exactly one solution in $[0, 1]$ (accounting for the fact that $t = \sin x$ maps $[0,\pi]$ to $[0,1]$ with $t=0$ giving one $x$ and $t \in (0,1)$ giving two $x$ values, and $t=1$ giving one $x$).
Step 4: For exactly one solution in $[0,\pi]$: either $t = 0$ is the only root in $[0,1]$ (giving $b = -1$ and $a$ arbitrary), or $t = 1$ is the only root in $[0,1]$, or one root is in $(0,1)$ and the other is outside $[0,1]$.
Step 5: Analyzing the conditions leads to $b \in (-\infty, -2] \cup (-1, \infty)$ and $a \neq b$. Options (2) and (4) are correct.
Correct Answer: 2, 4