To solve the inequality $\cos(x - \frac{\pi}{6}) \geq -\frac{1}{2}$, let's follow these steps:

Step 1: Solve the related equation

Start by solving the equation $\cos(x - \frac{\pi}{6}) = -\frac{1}{2}$.

We know that $\cos(\theta) = -\frac{1}{2}$ at specific angles, namely: $\theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \theta = \frac{4\pi}{3} + 2n\pi$ for any integer $n$.

Thus, for the equation $\cos(x - \frac{\pi}{6}) = -\frac{1}{2}$, we have: $x - \frac{\pi}{6} = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad x - \frac{\pi}{6} = \frac{4\pi}{3} + 2n\pi$

Step 2: Solve for $x$

Now solve these two equations for $x$:

1. $x - \frac{\pi}{6} = \frac{2\pi}{3} + 2n\pi$ $x = \frac{2\pi}{3} + \frac{\pi}{6} + 2n\pi = \frac{4\pi}{6} + \frac{\pi}{6} + 2n\pi = \frac{5\pi}{6} + 2n\pi$

2. $x - \frac{\pi}{6} = \frac{4\pi}{3} + 2n\pi$ $x = \frac{4\pi}{3} + \frac{\pi}{6} + 2n\pi = \frac{8\pi}{6} + \frac{\pi}{6} + 2n\pi = \frac{9\pi}{6} + 2n\pi = \frac{3\pi}{2} + 2n\pi$

Thus, the solutions to the equation $\cos(x - \frac{\pi}{6}) = -\frac{1}{2}$ are: $x = \frac{5\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{3\pi}{2} + 2n\pi$

Step 3: Analyze the inequality

The inequality $\cos(x - \frac{\pi}{6}) \geq -\frac{1}{2}$ means $x - \frac{\pi}{6}$ must be within the range of angles where $\cos(\theta)$ is greater than or equal to $-\frac{1}{2}$.

For $\cos(\theta) \geq -\frac{1}{2}$, the angles $\theta$ are within the range: $-\frac{\pi}{3} \leq \theta \leq \frac{5\pi}{3}$ This corresponds to the interval for $x - \frac{\pi}{6}$. So, we must solve: $-\frac{\pi}{3} \leq x - \frac{\pi}{6} \leq \frac{5\pi}{3}$

Step 4: Adjust for $x$

Add $\frac{\pi}{6}$ to each part of the inequality: $-\frac{\pi}{3} + \frac{\pi}{6} \leq x \leq \frac{5\pi}{3} + \frac{\pi}{6}$ Simplifying: $-\frac{2\pi}{6} + \frac{\pi}{6} = -\frac{\pi}{6}, \quad \frac{5\pi}{3} + \frac{\pi}{6} = \frac{10\pi}{6} + \frac{\pi}{6} = \frac{11\pi}{6}$

Thus, the inequality becomes: $-\frac{\pi}{6} \leq x \leq \frac{11\pi}{6}$

Step 5: General Solution

The general solution is the set of all $x$ values such that: $-\frac{\pi}{6} \leq x \leq \frac{11\pi}{6} + 2n\pi$ where $n$ is any integer, since the cosine function has a period of $2\pi$.

Final Answer:

The solution to the inequality $\cos(x - \frac{\pi}{6}) \geq -\frac{1}{2}$ is: $x \in \left[ -\frac{\pi}{6}, \frac{11\pi}{6} \right] + 2n\pi, \quad n \in \mathbb{Z}$

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Related Questions:

1. How would the solution change if the inequality was $\cos(x - \frac{\pi}{6}) \leq -\frac{1}{2}$?
2. How can you graph the function $y = \cos(x - \frac{\pi}{6})$ to visualize this inequality?
3. What other trigonometric inequalities are commonly solved in similar ways?
4. How can you apply this solution to real-world problems?
5. How does shifting the cosine function affect the solution to the inequality?

Tip:

When solving trigonometric inequalities, always consider the periodic nature of the functions and adjust the range of solutions accordingly by adding multiples of the period.