The given inequality is:

$\left|\sin(x) - \left(\frac{1}{2} + \frac{1}{\sqrt{2}}\right)\right| \geq \frac{1}{2} + \frac{1}{\sqrt{2}}$

Step-by-step solution:

1. Define constants:

Let’s first simplify the constant part $\frac{1}{2} + \frac{1}{\sqrt{2}}$. $\frac{1}{2} + \frac{1}{\sqrt{2}} \approx 0.5 + 0.707 \approx 1.207$

So, the inequality becomes: $|\sin(x) - 1.207| \geq 1.207$

2. Interpret the absolute value inequality:

The general inequality $|A - B| \geq C$ can be split into two cases: $\sin(x) - 1.207 \geq 1.207 \quad \text{or} \quad \sin(x) - 1.207 \leq -1.207$ which simplifies to: $\sin(x) \geq 1.207 + 1.207 = 2.414$ or $\sin(x) \leq 1.207 - 1.207 = 0$

3. Analyze the sine function’s range:

The sine function $\sin(x)$ has a range between $-1$ and $1$, i.e., $-1 \leq \sin(x) \leq 1$. However, $\sin(x) \geq 2.414$ is impossible because $2.414$ is outside the range of sine.

Thus, we are left with the inequality: $\sin(x) \leq 0$

4. Final answer:

The inequality holds true when $\sin(x) \leq 0$. This means $x$ must be in intervals where the sine function is non-positive, i.e., $x \in [\pi, 2\pi]$ (and other such intervals).

Conclusion:

The solution to the inequality is: $\sin(x) \leq 0$

This occurs in intervals like $[n\pi, (n+1)\pi]$, where $n$ is an integer.

Would you like more details or have any questions?

Here are some related questions:

1. What are the solutions of the inequality $\sin(x) \geq \frac{1}{2}$?
2. How would the inequality change if the constant was different?
3. Can you explain how to solve inequalities with $\cos(x)$ instead of $\sin(x)$?
4. How do we solve $\sin(x)$-based inequalities with absolute values?
5. What are some other common trigonometric inequalities?

Tip: When solving inequalities with absolute values, always consider both positive and negative cases to cover all possibilities.