We are given a system of equations (S): $x + y = \frac{\pi}{6}$ $sinx \cdot siny = -\frac{\sqrt{3}}{4}$ First, we need to show that $cos(x+y) - cos(x-y) = -2sinx \cdot siny$. Then, we need to show that system (S) is equivalent to system (S'): $x + y = \frac{\pi}{6}$ $cos(x-y) = 0$ Finally, we need to solve system (S).

1. Problem Description

We are given a system of equations (S):

$x + y = \frac{\pi}{6}$

$sinx \cdot siny = -\frac{\sqrt{3}}{4}$

First, we need to show that $cos(x+y) - cos(x-y) = -2sinx \cdot siny$. Then, we need to show that system (S) is equivalent to system (S'):

$x + y = \frac{\pi}{6}$

$cos(x-y) = 0$

Finally, we need to solve system (S).

2. Solution Steps

1.a. Prove that $cos(x+y) - cos(x-y) = -2sinx \cdot siny$

We use the following trigonometric identities:

$cos(a+b) = cosa \cdot cosb - sina \cdot sinb$

$cos(a-b) = cosa \cdot cosb + sina \cdot sinb$

Therefore,

$cos(x+y) - cos(x-y) = (cosx \cdot cosy - sinx \cdot siny) - (cosx \cdot cosy + sinx \cdot siny)$

$cos(x+y) - cos(x-y) = cosx \cdot cosy - sinx \cdot siny - cosx \cdot cosy - sinx \cdot siny$

$cos(x+y) - cos(x-y) = -2sinx \cdot siny$

1.b. Show that system (S) is equivalent to system (S'):

$x + y = \frac{\pi}{6}$

$sinx \cdot siny = -\frac{\sqrt{3}}{4}$

Since we have proven that $cos(x+y) - cos(x-y) = -2sinx \cdot siny$, we can substitute $sinx \cdot siny$ in the second equation of system (S):

$sinx \cdot siny = -\frac{\sqrt{3}}{4}$

$-2sinx \cdot siny = \frac{\sqrt{3}}{2}$

From the first equation, $x+y = \frac{\pi}{6}$. Therefore $cos(x+y) = cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.

Substituting $cos(x+y)$ into the equation $cos(x+y) - cos(x-y) = -2sinx \cdot siny$, we get:

$\frac{\sqrt{3}}{2} - cos(x-y) = \frac{\sqrt{3}}{2}$

$-cos(x-y) = 0$

$cos(x-y) = 0$

So, the equivalent system (S') is:

$x + y = \frac{\pi}{6}$

$cos(x-y) = 0$

2. Solve system (S):

The equivalent system (S') is:

$x + y = \frac{\pi}{6}$

$cos(x-y) = 0$

$x - y = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

Case 1: $x - y = \frac{\pi}{2}$

$x + y = \frac{\pi}{6}$

Adding the two equations, we get $2x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$

$x = \frac{\pi}{3}$.

$y = \frac{\pi}{6} - x = \frac{\pi}{6} - \frac{\pi}{3} = \frac{\pi - 2\pi}{6} = -\frac{\pi}{6}$

$x = \frac{\pi}{3}, y = -\frac{\pi}{6}$

Case 2: $x - y = -\frac{\pi}{2}$

$x + y = \frac{\pi}{6}$

Adding the two equations, we get $2x = -\frac{\pi}{2} + \frac{\pi}{6} = \frac{-\3\pi + \pi}{6} = \frac{-2\pi}{6} = -\frac{\pi}{3}$

$x = -\frac{\pi}{6}$

$y = \frac{\pi}{6} - x = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$

$x = -\frac{\pi}{6}, y = \frac{\pi}{3}$

3. Final Answer

The solutions to the system (S) are:

$(\frac{\pi}{3}, -\frac{\pi}{6})$ and $(-\frac{\pi}{6}, \frac{\pi}{3})$