If `[sin^(-1)x]^(2)+[sin^(-1)x]-2 le 0` (where, `[.]` represents the greatest integral part of x), then the maximum value of x is

Answer

To solve the inequality \([sin^{-1} x]^2 + [sin^{-1} x] - 2 \leq 0\), where \([.]\) represents the greatest integral part (floor function), we can follow these steps:

### Step 1: Let \( t = [sin^{-1} x] \)
We define \( t \) as the greatest integer less than or equal to \( sin^{-1} x \). Thus, we can rewrite the inequality as:
\[ t^2 + t - 2 \leq 0 \]

### Step 2: Factor the quadratic inequality
Next, we factor the quadratic expression:
\[ t^2 + t - 2 = (t + 2)(t - 1) \]
This gives us the inequality:
\[ (t + 2)(t - 1) \leq 0 \]

### Step 3: Determine the intervals for \( t \)
To solve the inequality \((t + 2)(t - 1) \leq 0\), we find the critical points, which are \( t = -2 \) and \( t = 1 \). We analyze the sign of the product in the intervals defined by these points:
- For \( t < -2 \): Both factors are negative, so the product is positive.
- For \( -2 \leq t \leq 1 \): The product is non-positive (either zero or negative).
- For \( t > 1 \): Both factors are positive, so the product is positive.

Thus, the solution to the inequality is:
\[ -2 \leq t \leq 1 \]

### Step 4: Find the maximum value of \( t \)
The maximum value of \( t \) in the interval \([-2, 1]\) is:
\[ t = 1 \]

### Step 5: Relate \( t \) back to \( sin^{-1} x \)
Since \( t = [sin^{-1} x] \), we have:
\[ [sin^{-1} x] = 1 \]
This means:
\[ 1 \leq sin^{-1} x < 2 \]

### Step 6: Solve for \( x \)
The range of \( sin^{-1} x \) is from \( 0 \) to \( \frac{\pi}{2} \) for \( x \) in \([0, 1]\). Therefore, we need to find the values of \( x \) such that:
\[ 1 \leq sin^{-1} x < 2 \]
Since \( sin^{-1} x \) can only reach a maximum of \( \frac{\pi}{2} \) (approximately \( 1.57 \)), we focus on the lower bound:
\[ sin^{-1} x \geq 1 \implies x \geq sin(1) \]

### Step 7: Calculate \( sin(1) \)
Using a calculator or known values:
\[ sin(1) \approx 0.8415 \]

### Conclusion
Thus, the maximum value of \( x \) that satisfies the original inequality is:
\[ \boxed{1} \]