The value of p for which the equation `3 " sin"^(2)x + 12 "cos" x - 3 =p` has at least one solution are

Options

A. `p le 12`
B. `3 le p le9`
C. `-15 le p le 9`
D. none of these

Explanation

To solve the equation \(3 \sin^2 x + 12 \cos x - 3 = p\) for the values of \(p\) for which the equation has at least one solution, we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ 3 \sin^2 x + 12 \cos x - 3 = p \]
Rearranging gives:
\[ 3 \sin^2 x + 12 \cos x - (p + 3) = 0 \]

### Step 2: Use the Pythagorean Identity
Using the identity \(\sin^2 x = 1 - \cos^2 x\), we can rewrite the equation in terms of \(\cos x\):
\[ 3(1 - \cos^2 x) + 12 \cos x - (p + 3) = 0 \]
This simplifies to:
\[ -3 \cos^2 x + 12 \cos x + (3 - p) = 0 \]
Multiplying through by -1 gives:
\[ 3 \cos^2 x - 12 \cos x + (p - 3) = 0 \]

### Step 3: Identify the Quadratic Form
This is a quadratic equation in \(\cos x\):
\[ 3 \cos^2 x - 12 \cos x + (p - 3) = 0 \]
Here, \(a = 3\), \(b = -12\), and \(c = p - 3\).

### Step 4: Determine Conditions for Real Solutions
For the quadratic equation to have at least one real solution, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Calculating the discriminant:
\[ D = (-12)^2 - 4 \cdot 3 \cdot (p - 3) \geq 0 \]
This simplifies to:
\[ 144 - 12(p - 3) \geq 0 \]
\[ 144 - 12p + 36 \geq 0 \]
\[ 180 - 12p \geq 0 \]

### Step 5: Solve for \(p\)
Rearranging gives:
\[ 12p \leq 180 \]
\[ p \leq 15 \]

### Step 6: Determine the Range of \(\cos x\)
Since \(\cos x\) ranges from -1 to 1, we need to ensure that the quadratic has solutions within this range. The roots of the quadratic can be found using:
\[ \cos x = \frac{12 \pm \sqrt{D}}{6} \]
We need:
\[ -1 \leq \frac{12 \pm \sqrt{D}}{6} \leq 1 \]

### Step 7: Solve the Inequalities
1. For the upper bound:
\[ \frac{12 + \sqrt{D}}{6} \leq 1 \]
This leads to:
\[ 12 + \sqrt{D} \leq 6 \implies \sqrt{D} \leq -6 \quad \text{(not possible)} \]

2. For the lower bound:
\[ \frac{12 - \sqrt{D}}{6} \geq -1 \]
This leads to:
\[ 12 - \sqrt{D} \geq -6 \implies \sqrt{D} \leq 18 \]

### Step 8: Combine Results
From the discriminant condition, we have \(p \leq 15\). Now, we also need to check the lower bound:
\[ D = 144 - 12(p - 3) \geq 0 \implies p \geq -15 \]

### Final Result
Thus, the values of \(p\) for which the equation has at least one solution are:
\[ -15 \leq p \leq 15 \]