The values of x which satisfy `18(sin^(-1)x)^(2)-9pi sin^(-1)x +pi^(2)lt 0` and `18(tan^(-1)x)^(2)-9pi tan^(-1)x + pi^(2)lt 0` simultaneously are

Options

A. `((sqrt(3))/(3),(sqrt(3))/(2))`
B. `((sqrt(3))/(2),1)`
C. `((1)/(3),(sqrt(3))/(2))`
D. `((1)/(sqrt(3)),sqrt(3))`

Explanation

To solve the inequalities \( 18(\sin^{-1} x)^2 - 9\pi \sin^{-1} x + \pi^2 < 0 \) and \( 18(\tan^{-1} x)^2 - 9\pi \tan^{-1} x + \pi^2 < 0 \) simultaneously, we can follow these steps:

### Step 1: Solve the first inequality
We start with the inequality:
\[ 18(\sin^{-1} x)^2 - 9\pi \sin^{-1} x + \pi^2 < 0 \]
Divide the entire inequality by 18:
\[ (\sin^{-1} x)^2 - \frac{9\pi}{18} \sin^{-1} x + \frac{\pi^2}{18} < 0 \]
This simplifies to:
\[ (\sin^{-1} x)^2 - \frac{\pi}{2} \sin^{-1} x + \frac{\pi^2}{18} < 0 \]

### Step 2: Factor the quadratic
Next, we need to factor the quadratic:
\[ (\sin^{-1} x)^2 - \frac{\pi}{2} \sin^{-1} x + \frac{\pi^2}{18} < 0 \]
We can find the roots using the quadratic formula:
\[ \sin^{-1} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -\frac{\pi}{2}, c = \frac{\pi^2}{18} \):
\[ \sin^{-1} x = \frac{\frac{\pi}{2} \pm \sqrt{\left(-\frac{\pi}{2}\right)^2 - 4 \cdot 1 \cdot \frac{\pi^2}{18}}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \left(-\frac{\pi}{2}\right)^2 - 4 \cdot 1 \cdot \frac{\pi^2}{18} = \frac{\pi^2}{4} - \frac{4\pi^2}{18} = \frac{\pi^2}{4} - \frac{2\pi^2}{9} \]
Finding a common denominator (36):
\[ \frac{9\pi^2}{36} - \frac{8\pi^2}{36} = \frac{\pi^2}{36} \]
Now substituting back:
\[ \sin^{-1} x = \frac{\frac{\pi}{2} \pm \frac{\pi}{6}}{2} \]
Calculating the roots:
\[ \sin^{-1} x = \frac{3\pi}{12} = \frac{\pi}{4} \quad \text{and} \quad \sin^{-1} x = \frac{\pi}{12} \]

### Step 3: Determine the intervals
The roots are \( \frac{\pi}{12} \) and \( \frac{\pi}{4} \). The quadratic opens upwards, so the inequality \( < 0 \) holds between the roots:
\[ \frac{\pi}{12} < \sin^{-1} x < \frac{\pi}{4} \]
Taking the sine of all parts:
\[ \sin\left(\frac{\pi}{12}\right) < x < \sin\left(\frac{\pi}{4}\right) \]
Calculating these values:
\[ \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]

### Step 4: Solve the second inequality
Now we solve the second inequality:
\[ 18(\tan^{-1} x)^2 - 9\pi \tan^{-1} x + \pi^2 < 0 \]
Following the same steps as before, we find:
\[ (\tan^{-1} x)^2 - \frac{\pi}{2} \tan^{-1} x + \frac{\pi^2}{18} < 0 \]
The roots will be the same as for the sine case, so:
\[ \frac{\pi}{12} < \tan^{-1} x < \frac{\pi}{4} \]
Taking the tangent of all parts:
\[ \tan\left(\frac{\pi}{12}\right) < x < \tan\left(\frac{\pi}{4}\right) \]
Calculating these values:
\[ \tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 5: Combine the intervals
Now we combine the intervals from both inequalities:
1. From the sine inequality:
\[ \frac{\sqrt{6} - \sqrt{2}}{4} < x < \frac{\sqrt{2}}{2} \]
2. From the tangent inequality:
\[ 2 - \sqrt{3} < x < 1 \]

### Final Result
The values of \( x \) that satisfy both inequalities simultaneously are:
\[ \max\left(\frac{\sqrt{6} - \sqrt{2}}{4}, 2 - \sqrt{3}\right) < x < \min\left(\frac{\sqrt{2}}{2}, 1\right) \]