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

1. Problem Description

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

We know that $\sin(\frac{5\pi}{4}) = - \frac{\sqrt{2}}{2}$ and $\sin(\frac{7\pi}{4}) = - \frac{\sqrt{2}}{2}$. Thus,

$x - \frac{\pi}{6} = \frac{5\pi}{4} + 2n\pi$ or $x - \frac{\pi}{6} = \frac{7\pi}{4} + 2n\pi$, where $n$ is an integer.

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

Then $x = \frac{5\pi}{4} + \frac{\pi}{6} + 2n\pi = \frac{15\pi}{12} + \frac{2\pi}{12} + 2n\pi = \frac{17\pi}{12} + 2n\pi$.

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

Then $x = \frac{7\pi}{4} + \frac{\pi}{6} + 2n\pi = \frac{21\pi}{12} + \frac{2\pi}{12} + 2n\pi = \frac{23\pi}{12} + 2n\pi$.

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

If we are looking for solutions in the interval $[0, 2\pi)$, then for $n=0$, we have $x = \frac{17\pi}{12}$ and $x = \frac{23\pi}{12}$.

3. Final Answer

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