Solve the trigonometric equation $\sin(x - \frac{\pi}{6}) = -\frac{\sqrt{2}}{2}$ for $x$.

1. Problem Description

Solve the trigonometric equation $\sin(x - \frac{\pi}{6}) = -\frac{\sqrt{2}}{2}$ for $x$.

2. Solution Steps

We want to find the values of $x$ that satisfy the equation $\sin(x - \frac{\pi}{6}) = -\frac{\sqrt{2}}{2}$.

First, we find the angles whose sine is $-\frac{\sqrt{2}}{2}$. We know that $\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$ and $\sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2}$.

Therefore, the general solutions for the sine function are given by:

$x - \frac{\pi}{6} = \frac{5\pi}{4} + 2n\pi$

$x - \frac{\pi}{6} = \frac{7\pi}{4} + 2n\pi$

where $n$ is an integer.

Now, we solve for $x$ in each case:

Case 1: $x - \frac{\pi}{6} = \frac{5\pi}{4} + 2n\pi$

$x = \frac{5\pi}{4} + \frac{\pi}{6} + 2n\pi$

$x = \frac{15\pi + 2\pi}{12} + 2n\pi$

$x = \frac{17\pi}{12} + 2n\pi$

Case 2: $x - \frac{\pi}{6} = \frac{7\pi}{4} + 2n\pi$

$x = \frac{7\pi}{4} + \frac{\pi}{6} + 2n\pi$

$x = \frac{21\pi + 2\pi}{12} + 2n\pi$

$x = \frac{23\pi}{12} + 2n\pi$

The solutions are $x = \frac{17\pi}{12} + 2n\pi$ and $x = \frac{23\pi}{12} + 2n\pi$, where $n$ is an integer.

3. Final Answer

$x = \frac{17\pi}{12} + 2n\pi, \frac{23\pi}{12} + 2n\pi$