Suppose ` 3 sin ^(-1) ( log _(2) x ) + cos^(-1) ( log _(2) y) =pi //2 " and " sin^(-1) ( log _(2) x ) + 2 cos^(-1) ( log_(2) y) = 11 pi//6`, then the value of ` x^(-2) + y^(-2)` equals

Answer

To solve the given problem, we start with the two equations provided:

1. \( 3 \sin^{-1}(\log_2 x) + \cos^{-1}(\log_2 y) = \frac{\pi}{2} \)
2. \( \sin^{-1}(\log_2 x) + 2 \cos^{-1}(\log_2 y) = \frac{11\pi}{6} \)

Let's denote:
- \( p = \sin^{-1}(\log_2 x) \)
- \( q = \cos^{-1}(\log_2 y) \)

Now we can rewrite the equations in terms of \( p \) and \( q \):

1. \( 3p + q = \frac{\pi}{2} \) (Equation 1)
2. \( p + 2q = \frac{11\pi}{6} \) (Equation 2)

### Step 1: Solve for \( p \) and \( q \)

From Equation 1, we can express \( q \) in terms of \( p \):

\[ q = \frac{\pi}{2} - 3p \]

### Step 2: Substitute \( q \) into Equation 2

Now substitute \( q \) into Equation 2:

\[ p + 2\left(\frac{\pi}{2} - 3p\right) = \frac{11\pi}{6} \]

### Step 3: Simplify the equation

Expanding this gives:

\[ p + \pi - 6p = \frac{11\pi}{6} \]

Combining like terms results in:

\[ -5p + \pi = \frac{11\pi}{6} \]

### Step 4: Isolate \( p \)

To isolate \( p \), we can rearrange the equation:

\[ -5p = \frac{11\pi}{6} - \pi \]

Converting \( \pi \) to sixths:

\[ -5p = \frac{11\pi}{6} - \frac{6\pi}{6} = \frac{5\pi}{6} \]

Dividing both sides by -5 gives:

\[ p = -\frac{\pi}{6} \]

### Step 5: Find \( q \)

Now substitute \( p \) back into the expression for \( q \):

\[ q = \frac{\pi}{2} - 3\left(-\frac{\pi}{6}\right) \]

Calculating gives:

\[ q = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Step 6: Find \( \log_2 x \) and \( \log_2 y \)

Now we have:

- \( p = \sin^{-1}(\log_2 x) = -\frac{\pi}{6} \)
- \( q = \cos^{-1}(\log_2 y) = \pi \)

From \( p \):

\[ \log_2 x = \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \]
Thus,

\[ x = 2^{-\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

From \( q \):

\[ \log_2 y = \cos(\pi) = -1 \]
Thus,

\[ y = 2^{-1} = \frac{1}{2} \]

### Step 7: Calculate \( x^{-2} + y^{-2} \)

Now we need to find:

\[ x^{-2} + y^{-2} \]

Calculating \( x^{-2} \) and \( y^{-2} \):

\[ x^{-2} = \left(\frac{1}{\sqrt{2}}\right)^{-2} = 2 \]
\[ y^{-2} = \left(\frac{1}{2}\right)^{-2} = 4 \]

Thus,

\[ x^{-2} + y^{-2} = 2 + 4 = 6 \]

### Final Answer

The value of \( x^{-2} + y^{-2} \) is \( \boxed{6} \).

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