Given `12 sin theta + 5 cos theta =2 x^(2) - 8 x +21` and `theta ` and x are the solutions of above then `theta` x is

Options

A. `pi -2 tan^(-1) ""( 5)/( 12)`
B. `pi -2 tan^(-1) ""( 12)/( 5)`
C. 0
D. none

Answer

To solve the equation \( 12 \sin \theta + 5 \cos \theta = 2x^2 - 8x + 21 \) for \( \theta \) and \( x \), we can follow these steps:

### Step 1: Rewrite the left-hand side
We can express \( 12 \sin \theta + 5 \cos \theta \) in a different form. We know that it can be rewritten using the sine addition formula.

Let \( R = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \).

Now, we can express \( 12 \sin \theta + 5 \cos \theta \) as:
\[ R (\sin \theta \cos \alpha + \cos \theta \sin \alpha) = R \sin(\theta + \alpha) \]
where \( \cos \alpha = \frac{12}{13} \) and \( \sin \alpha = \frac{5}{13} \).

### Step 2: Set up the equation
Thus, we can rewrite the equation as:
\[ 13 \sin(\theta + \alpha) = 2x^2 - 8x + 21 \]

### Step 3: Determine the ranges
The maximum value of \( 13 \sin(\theta + \alpha) \) is \( 13 \) and the minimum value is \( -13 \). Therefore, we have:
\[ -13 \leq 2x^2 - 8x + 21 \leq 13 \]

### Step 4: Solve the inequalities
1. For the upper bound:
\[ 2x^2 - 8x + 21 \leq 13 \]
Simplifying gives:
\[ 2x^2 - 8x + 8 \leq 0 \implies x^2 - 4x + 4 \leq 0 \implies (x - 2)^2 \leq 0 \]
This implies \( x = 2 \).

2. For the lower bound:
\[ 2x^2 - 8x + 21 \geq -13 \]
Simplifying gives:
\[ 2x^2 - 8x + 34 \geq 0 \]
This quadratic has no real roots (discriminant \( (-8)^2 - 4 \cdot 2 \cdot 34 < 0 \)), meaning it is always positive.

### Step 5: Find \( \theta \)
Since \( x = 2 \) is the only solution, we substitute \( x = 2 \) back into the original equation:
\[ 12 \sin \theta + 5 \cos \theta = 2(2^2) - 8(2) + 21 \]
Calculating the right-hand side:
\[ = 8 - 16 + 21 = 13 \]
Thus, we have:
\[ 12 \sin \theta + 5 \cos \theta = 13 \]
Using the earlier transformation:
\[ 13 \sin(\theta + \alpha) = 13 \]
This implies:
\[ \sin(\theta + \alpha) = 1 \implies \theta + \alpha = \frac{\pi}{2} \implies \theta = \frac{\pi}{2} - \alpha \]

### Step 6: Find \( \alpha \)
From \( \cos \alpha = \frac{12}{13} \) and \( \sin \alpha = \frac{5}{13} \):
\[ \alpha = \sin^{-1}\left(\frac{5}{13}\right) \]

### Step 7: Calculate \( \theta x \)
Now we need to find \( \theta x \):
\[ \theta = \frac{\pi}{2} - \sin^{-1}\left(\frac{5}{13}\right) \]
Thus:
\[ \theta x = \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{5}{13}\right)\right) \cdot 2 \]
This simplifies to:
\[ \theta x = \pi - 2 \sin^{-1}\left(\frac{5}{13}\right) \]

### Final Answer
\[ \theta x = \pi - 2 \tan^{-1}\left(\frac{5}{12}\right) \]