If `|sin^2 x + 17 - x ^2| = |16 - x^2| + 2sin^2 x + cos^2 x` then subsets of solution are

Explanation

To solve the equation \( | \sin^2 x + 17 - x^2 | = | 16 - x^2 | + 2 \sin^2 x + \cos^2 x \), we can follow these steps:

### Step 1: Simplify the Right-Hand Side
We know that \( \sin^2 x + \cos^2 x = 1 \). Therefore, we can rewrite the right-hand side of the equation:

\[ |16 - x^2| + 2 \sin^2 x + \cos^2 x = |16 - x^2| + 2 \sin^2 x + 1 \]

### Step 2: Rewrite the Equation
Now, substitute this back into the original equation:

\[ | \sin^2 x + 17 - x^2 | = |16 - x^2| + 2 \sin^2 x + 1 \]

### Step 3: Analyze the Absolute Values
We need to consider the cases for the absolute values.

**Case 1:** When \( \sin^2 x + 17 - x^2 \geq 0 \) and \( 16 - x^2 \geq 0 \)

In this case, we can drop the absolute values:

\[ \sin^2 x + 17 - x^2 = 16 - x^2 + 2 \sin^2 x + 1 \]

Simplifying this gives:

\[ \sin^2 x + 17 - x^2 = 17 - x^2 + 2 \sin^2 x \]

Rearranging terms:

\[ \sin^2 x = 2 \sin^2 x \]

This leads to:

\[ \sin^2 x = 0 \implies \sin x = 0 \]

The solutions for \( \sin x = 0 \) are:

\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]

**Case 2:** When \( \sin^2 x + 17 - x^2 < 0 \) and \( 16 - x^2 \geq 0 \)

In this case, we have:

\[ -(\sin^2 x + 17 - x^2) = 16 - x^2 + 2 \sin^2 x + 1 \]

This simplifies to:

\[ -x^2 - \sin^2 x - 17 = 16 - x^2 + 2 \sin^2 x + 1 \]

Rearranging gives:

\[ -17 = 16 + 1 + 3 \sin^2 x \]

This leads to:

\[ -17 = 17 + 3 \sin^2 x \implies 3 \sin^2 x = -34 \quad \text{(not possible)} \]

Thus, there are no solutions from this case.

**Case 3:** When \( \sin^2 x + 17 - x^2 \geq 0 \) and \( 16 - x^2 < 0 \)

In this case, we have:

\[ \sin^2 x + 17 - x^2 = -(16 - x^2) + 2 \sin^2 x + 1 \]

This simplifies to:

\[ \sin^2 x + 17 - x^2 = -16 + x^2 + 2 \sin^2 x + 1 \]

Rearranging gives:

\[ \sin^2 x + 17 - x^2 = -15 + x^2 + 2 \sin^2 x \]

This leads to:

\[ -2x^2 + \sin^2 x + 32 = 0 \]

This is a quadratic equation in \( \sin^2 x \).

### Step 4: Solve for \( x \)
After analyzing the cases, we find that valid solutions occur when \( x = 4 \) or \( x = -4 \).

### Final Solution
Thus, the subsets of solutions are:

\[ x \in \{-4, 4\} \]