Solve for x and y `12sinx-2y^2=21-8y-5 cosx`

Explanation

To solve the equation \( 12\sin x - 2y^2 = 21 - 8y - 5\cos x \), we will rearrange and analyze the equation step by step.

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

### Step 2: Analyzing the Left-Hand Side (LHS)
The LHS is \( 12\sin x + 5\cos x \). We can express this in the form \( R\sin(x + \alpha) \) where:
\[ R = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]
To find \( \tan \alpha \):
\[ \tan \alpha = \frac{5}{12} \]
Thus, we can write:
\[ 12\sin x + 5\cos x = 13\sin(x + \alpha) \]
where \( \alpha = \tan^{-1}\left(\frac{5}{12}\right) \).

### Step 3: Finding the Range of LHS
Since \( \sin(x + \alpha) \) ranges from -1 to 1, we have:
\[ -13 \leq 12\sin x + 5\cos x \leq 13 \]

### Step 4: Analyzing the Right-Hand Side (RHS)
The RHS is \( 2y^2 - 8y + 21 \). We can rewrite this as:
\[ 2(y^2 - 4y) + 21 \]
Completing the square:
\[ = 2((y - 2)^2 - 4) + 21 = 2(y - 2)^2 + 13 \]
This shows that:
\[ 2(y - 2)^2 \geq 0 \implies 2(y - 2)^2 + 13 \geq 13 \]
Thus, the minimum value of the RHS is 13.

### Step 5: Setting LHS Equal to RHS
Since the LHS can be at most 13 and the RHS is at least 13, we set:
\[ 12\sin x + 5\cos x = 13 \]
This implies:
\[ \sin(x + \alpha) = 1 \]
Thus:
\[ x + \alpha = \frac{\pi}{2} + 2k\pi \quad (k \in \mathbb{Z}) \]
This leads to:
\[ x = \frac{\pi}{2} - \alpha + 2k\pi \]

### Step 6: Solving for y
From the RHS, we have:
\[ 2(y - 2)^2 + 13 = 13 \implies 2(y - 2)^2 = 0 \implies y - 2 = 0 \implies y = 2 \]

### Final Solutions
Thus, we find:
\[ y = 2 \]
And substituting for \( \alpha \):
\[ \alpha = \tan^{-1}\left(\frac{5}{12}\right) \]
So:
\[ x = \frac{\pi}{2} - \tan^{-1}\left(\frac{5}{12}\right) + 2k\pi \]
Alternatively, using the cotangent identity:
\[ x = \cot^{-1}\left(\frac{5}{12}\right) + 2k\pi \]

### Summary of Solutions
The solutions are:
\[ y = 2 \]
\[ x = \cot^{-1}\left(\frac{5}{12}\right) + 2k\pi \quad (k \in \mathbb{Z}) \]