If `|sin^(2)x+10x^(2)|=|9-x^(2)|+2sin^(2)x+cos^(2)x`, then x lies in

Options

A. `[-8. 8]`
B. `[-3, 3]`
C. `[-sqrt(17), sqrt(17)]`
D. `[-sqrt(21), sqrt(21)]`

Answer

To solve the equation \( |\sin^2 x + 10x^2| = |9 - x^2| + 2\sin^2 x + \cos^2 x \), we will follow these steps:

### Step 1: Simplify the Equation
Start by using the identity \( \sin^2 x + \cos^2 x = 1 \) to rewrite the equation:
\[ |\sin^2 x + 10x^2| = |9 - x^2| + 2\sin^2 x + \cos^2 x \]
Substituting \( \cos^2 x = 1 - \sin^2 x \):
\[ |\sin^2 x + 10x^2| = |9 - x^2| + 2\sin^2 x + (1 - \sin^2 x) \]
This simplifies to:
\[ |\sin^2 x + 10x^2| = |9 - x^2| + \sin^2 x + 1 \]

### Step 2: Rearranging the Equation
Rearranging gives:
\[ |\sin^2 x + 10x^2| - \sin^2 x - 1 = |9 - x^2| \]

### Step 3: Analyze the Absolute Values
Now we need to consider the cases for the absolute values. We have two absolute values to analyze: \( |\sin^2 x + 10x^2| \) and \( |9 - x^2| \).

1. **Case 1**: \( \sin^2 x + 10x^2 \geq 0 \)
- Then \( |\sin^2 x + 10x^2| = \sin^2 x + 10x^2 \).
- The equation becomes:
\[ \sin^2 x + 10x^2 - \sin^2 x - 1 = 9 - x^2 \]
Simplifying gives:
\[ 10x^2 - 1 = 9 - x^2 \]
Rearranging leads to:
\[ 11x^2 = 10 \implies x^2 = \frac{10}{11} \]

2. **Case 2**: \( \sin^2 x + 10x^2 < 0 \)
- Then \( |\sin^2 x + 10x^2| = -(\sin^2 x + 10x^2) \).
- The equation becomes:
\[ -(\sin^2 x + 10x^2) - \sin^2 x - 1 = 9 - x^2 \]
Simplifying gives:
\[ -2\sin^2 x - 10x^2 - 1 = 9 - x^2 \]
Rearranging leads to:
\[ -2\sin^2 x - 9x^2 = 10 \implies 2\sin^2 x + 9x^2 = -10 \]
This case will not yield any valid solutions since \( \sin^2 x \) and \( x^2 \) are always non-negative.

### Step 4: Determine Valid Range for \( x \)
From the first case, we found \( x^2 = \frac{10}{11} \). We also need to ensure that \( 9 - x^2 \geq 0 \):
\[ 9 - x^2 \geq 0 \implies x^2 \leq 9 \implies -3 \leq x \leq 3 \]

### Conclusion
Combining the results, we find that \( x \) must satisfy:
\[ -\sqrt{9} \leq x \leq \sqrt{9} \implies -3 \leq x \leq 3 \]

Thus, the final answer is:
\[ \text{The values of } x \text{ lie in } [-3, 3]. \]