Find Sum of Integral Values of p for Solvability of Trigonometric Equation

To find the sum of all integral values of $p$ for which the equation $3\sin^2 x + 12\cos x - 3 = p, \quad x \in \mathbb{R}$ has at least one real solution, we start by expressing the equation entirely in terms of $\cos x$.

Using the identity $\sin^2 x = 1 - \cos^2 x$, we substitute it into the given equation: $3(1 - \cos^2 x) + 12\cos x - 3 = p$

Simplifying the left-hand side: $3 - 3\cos^2 x + 12\cos x - 3 = p$ $-3\cos^2 x + 12\cos x = p$

Let $t = \cos x$. Since $x \in \mathbb{R}$, the variable $t$ lies in the interval $[-1, 1]$. Thus, the equation reduces to finding the range of the function: $f(t) = -3t^2 + 12t \quad \text{for } t \in [-1, 1]$

To determine the extrema of $f(t)$, we find its derivative with respect to $t$: $f'(t) = -6t + 12 = 6(2 - t)$

For all $t \in [-1, 1]$, $2 - t > 0$, which implies $f'(t) > 0$. Hence, $f(t)$ is a strictly increasing function on the closed interval $[-1, 1]$.

• The minimum value of $f(t)$ occurs at $t = -1$: $f(-1) = -3(-1)^2 + 12(-1) = -3 - 12 = -15$

• The maximum value of $f(t)$ occurs at $t = 1$: $f(1) = -3(1)^2 + 12(1) = -3 + 12 = 9$

Therefore, the range of $f(t)$ for $t \in [-1, 1]$ is $[-15, 9]$.

For the given trigonometric equation to have at least one real solution, $p$ must belong to this range: $p \in [-15, 9]$

Now, we calculate the sum of all integral values of $p$ in the interval $[-15, 9]$: $\text{Sum} = \sum_{p=-15}^{9} p$

By splitting the summation: $\text{Sum} = \sum_{p=-15}^{-10} p + \sum_{p=-9}^{9} p$

Since the sum of integers from $-9$ to $9$ is $0$ ($\sum_{p=-9}^{9} p = 0$), the total sum reduces to: $\text{Sum} = (-15) + (-14) + (-13) + (-12) + (-11) + (-10)$ $\text{Sum} = -(15 + 14 + 13 + 12 + 11 + 10) = -75$

Hence, the correct option is C.