\left|\sin\left(x\right)-\left(\frac{1}{2}+\frac{1}{\sqrt[2]{2}}\right)\right|\ge\frac{1}{2}+\frac{1}{\sqrt[2]{2}}

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.

The query submitted by the user: |sin(x) - (1/2 + 1/√2)| ≥ 1/2 + 1/√2

Discover how to solve the trigonometric inequality |sin(x) - (1/2 + 1/√2)| ≥ 1/2 + 1/√2 with a step-by-step breakdown. Learn about the absolute value inequality and the properties of the sine function. This detailed explanation is ideal for high school students in Grades 9-12.

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