If `x=2npi+tan^(-1).(p)/(q) and y=r` is a solution of the equation
`12sinx+5cosx=2y^(2)-8y+21`, then the value of k, such that `sqrt(p^(2)+q^(2)+kr^(2))=15`, is equal to

Options

A. 5
B. 14
C. `(31)/(4)`
D. `-22`

Answer

To solve the problem step-by-step, we will follow the logical flow of the video transcript while providing a clear mathematical approach.

### Step 1: Analyze the given equation
We start with the equation:
\[ 12 \sin x + 5 \cos x = 2y^2 - 8y + 21 \]

### Step 2: Identify the left-hand side (LHS)
The left-hand side can be expressed in a different form. We can rewrite it using the formula for a linear combination of sine and cosine:
\[ LHS = 12 \sin x + 5 \cos x \]

### Step 3: Find the amplitude and phase shift
We can express \( LHS \) as:
\[ LHS = a \sin(x + \alpha) \]
where \( a = \sqrt{12^2 + 5^2} \) and \( \tan \alpha = \frac{5}{12} \).

Calculating \( a \):
\[ a = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

### Step 4: Determine the angle \( \alpha \)
Using the tangent:
\[ \tan \alpha = \frac{5}{12} \implies \alpha = \tan^{-1}\left(\frac{5}{12}\right) \]

### Step 5: Rewrite LHS using the sine addition formula
Now we can rewrite the LHS:
\[ LHS = 13 \sin(x + \alpha) \]

### Step 6: Establish the maximum value of LHS
The maximum value of \( \sin(x + \alpha) \) is 1, thus:
\[ LHS \leq 13 \]

### Step 7: Analyze the right-hand side (RHS)
The right-hand side is a quadratic in \( y \):
\[ RHS = 2y^2 - 8y + 21 \]

### Step 8: Set up the inequality
Since \( LHS \leq 13 \), we set the RHS to be less than or equal to 13:
\[ 2y^2 - 8y + 21 \leq 13 \]
This simplifies to:
\[ 2y^2 - 8y + 8 \leq 0 \]
Dividing through by 2 gives:
\[ y^2 - 4y + 4 \leq 0 \]

### Step 9: Factor the quadratic
Factoring gives:
\[ (y - 2)^2 \leq 0 \]
This implies:
\[ y - 2 = 0 \implies y = 2 \]

### Step 10: Substitute \( y \) back into the equation
Since \( y = r \), we have:
\[ r = 2 \]

### Step 11: Substitute \( y \) into the original equation
Now substituting \( y = 2 \) back into the equation:
\[ 12 \sin x + 5 \cos x = 2(2^2) - 8(2) + 21 \]
Calculating the RHS:
\[ = 8 - 16 + 21 = 13 \]
Thus:
\[ 12 \sin x + 5 \cos x = 13 \]

### Step 12: Set the equation for \( x \)
From our previous analysis, we have:
\[ 13 \sin(x + \alpha) = 13 \implies \sin(x + \alpha) = 1 \]
This means:
\[ x + \alpha = \frac{\pi}{2} + 2n\pi \implies x = \frac{\pi}{2} - \alpha + 2n\pi \]

### Step 13: Solve for \( p \) and \( q \)
Since \( \alpha = \tan^{-1}\left(\frac{5}{12}\right) \), we can express \( x \) as:
\[ x = 2n\pi + \tan^{-1}\left(\frac{p}{q}\right) \]
Comparing gives \( p = 5 \) and \( q = 12 \).

### Step 14: Set up the equation for \( k \)
Now we need to find \( k \) such that:
\[ \sqrt{p^2 + q^2 + kr^2} = 15 \]
Squaring both sides:
\[ p^2 + q^2 + kr^2 = 225 \]

### Step 15: Substitute known values
Substituting \( p = 5 \), \( q = 12 \), and \( r = 2 \):
\[ 5^2 + 12^2 + k(2^2) = 225 \]
Calculating:
\[ 25 + 144 + 4k = 225 \]
This simplifies to:
\[ 169 + 4k = 225 \]
Thus:
\[ 4k = 225 - 169 = 56 \implies k = \frac{56}{4} = 14 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{14} \]