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 will follow these steps:

### Step 1: Let \( t = [ \sin^{-1} x ] \)
We can rewrite the inequality as:
\[ t^2 + t - 2 \leq 0 \]

### Step 2: Factor the quadratic expression
To factor the quadratic, we look for two numbers that multiply to \(-2\) and add to \(1\). The factors are:
\[ (t - 1)(t + 2) \leq 0 \]

### Step 3: Determine the intervals for \( t \)
The roots of the equation \( (t - 1)(t + 2) = 0 \) are \( t = 1 \) and \( t = -2 \). We need to find the intervals where the product is less than or equal to zero:
- The intervals are: \( (-\infty, -2] \), \( [-2, 1] \), and \( [1, \infty) \).
- Testing these intervals:
- For \( t < -2 \) (e.g., \( t = -3 \)): \( (-3 - 1)(-3 + 2) = (-4)(-1) > 0 \)
- For \( -2 < t < 1 \) (e.g., \( t = 0 \)): \( (0 - 1)(0 + 2) = (-1)(2) < 0 \)
- For \( t > 1 \) (e.g., \( t = 2 \)): \( (2 - 1)(2 + 2) = (1)(4) > 0 \)

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

### Step 4: Interpret the values of \( t \)
Since \( t = [ \sin^{-1} x ] \), we need to find the possible values of \( \sin^{-1} x \):
- The greatest integral part \( t \) can take values in the range \( -2 \) to \( 1 \).

### Step 5: Determine the range of \( \sin^{-1} x \)
The function \( \sin^{-1} x \) has a range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\) or approximately \([-1.57, 1.57]\). Thus:
- If \( t = -2 \), then \( \sin^{-1} x \) can be any value that gives a floor of \(-2\), which is not possible since the maximum value of \(\sin^{-1} x\) is \(1\).
- If \( t = -1\), then \(-1 < \sin^{-1} x < 0\) which corresponds to \( -1 < x < 0 \).
- If \( t = 0\), then \( 0 \leq \sin^{-1} x < 1\) which corresponds to \( 0 \leq x < \frac{\sqrt{2}}{2} \).
- If \( t = 1\), then \( 1 \leq \sin^{-1} x < 2\) which corresponds to \( x = 1 \).

### Step 6: Find the maximum value of \( x \)
The maximum value of \( x \) occurs when \( t = 1\), which gives:
\[ \sin^{-1} x = 1 \implies x = \sin(1) \approx 1 \]

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