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The problem asks to calculate the values of \(x\) and \(y\) given the system of equations:

\[
\text{sin}y = x^2 + \text{sin}x, \quad \text{cos}x = y^2 + \text{cos}y
\]

where \(x = \frac{\pi}{6}\).

Let's solve this system of equations.

### Step 1: Substitute \(x = \frac{\pi}{6}\) into the equations

Substitute \(x = \frac{\pi}{6}\) into the first equation:

\[
\text{sin}y = \left(\frac{\pi}{6}\right)^2 + \text{sin}\left(\frac{\pi}{6}\right)
\]

We know that:

\[
\text{sin}\left(\frac{\pi}{6}\right) = \frac{1}{2}
\]

Thus, the equation becomes:

\[
\text{sin}y = \left(\frac{\pi}{6}\right)^2 + \frac{1}{2}
\]

Next, let's calculate \(\left(\frac{\pi}{6}\right)^2\):

\[
\left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{36}
\]

Thus, the equation becomes:

\[
\text{sin}y = \frac{\pi^2}{36} + \frac{1}{2}
\]

### Step 2: Substitute \(x = \frac{\pi}{6}\) into the second equation

Substitute \(x = \frac{\pi}{6}\) into the second equation:

\[
\text{cos}\left(\frac{\pi}{6}\right) = y^2 + \text{cos}y
\]

We know that:

\[
\text{cos}\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
\]

Thus, the equation becomes:

\[
\frac{\sqrt{3}}{2} = y^2 + \text{cos}y
\]

### Step 3: Solve the system of equations

We have two equations to solve:

1. \(\text{sin}y = \frac{\pi^2}{36} + \frac{1}{2}\)
2. \(\frac{\sqrt{3}}{2} = y^2 + \text{cos}y\)

Let's solve these numerically using approximation methods or by substituting potential values for \(y\) and checking which satisfy both equations. I'll perform these calculations.

Learn how to solve the system of trigonometric equations involving sine and cosine functions to find the values of x and y. This problem requires understanding trigonometric identities and involves numerical methods for approximation. Suitable for advanced high school or college-level students.

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