The solution of `sin^(-1)|sin x|=sqrt(sin^(-1)|sin x|)` is

Options

A. `n pi pm 1, n pi, n in Z`
B. `n pi+1, n pi, n in Z`
C. `n pi-1, n pi, n in Z`
D. `2n pi+1, n pi, n in Z`

Explanation

To solve the equation \( \sin^{-1}|\sin x| = \sqrt{\sin^{-1}|\sin x|} \), we will follow these steps:

### Step 1: Let \( y = \sin^{-1}|\sin x| \)

We start by substituting \( y \) for \( \sin^{-1}|\sin x| \). This gives us the equation:

\[ y = \sqrt{y} \]

### Step 2: Square both sides

Next, we square both sides of the equation to eliminate the square root:

\[ y^2 = y \]

### Step 3: Rearrange the equation

Rearranging the equation gives us:

\[ y^2 - y = 0 \]

### Step 4: Factor the equation

Factoring the equation results in:

\[ y(y - 1) = 0 \]

### Step 5: Solve for \( y \)

Setting each factor to zero gives us the solutions:

\[ y = 0 \quad \text{or} \quad y = 1 \]

### Step 6: Substitute back to find \( x \)

Now we substitute back to find \( x \) using \( y = \sin^{-1}|\sin x| \).

1. For \( y = 0 \):
\[ \sin^{-1}|\sin x| = 0 \implies |\sin x| = 0 \implies \sin x = 0 \]
The solutions for \( \sin x = 0 \) are:
\[ x = n\pi, \quad n \in \mathbb{Z} \]

2. For \( y = 1 \):
\[ \sin^{-1}|\sin x| = 1 \implies |\sin x| = \sin(1) \]
The solutions for \( |\sin x| = \sin(1) \) are:
\[ x = n\pi + (-1)^k \cdot 1, \quad n \in \mathbb{Z}, \quad k = 0, 1 \]

### Final Solution

Combining both sets of solutions, we have:

\[ x = n\pi \quad \text{or} \quad x = n\pi + 1 \quad \text{or} \quad x = n\pi - 1, \quad n \in \mathbb{Z} \]