How would you solve the inequality sin (2x) 2 sin x and what is the answer to the inequality?
How would you solve the inequality sin (2x) 2 sin x and what is the answer to the inequality?
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
How would you solve the inequality sin (2x) ≥ sin x and what is the answer to the inequality?
![**How would you solve the inequality sin(2x) ≥ sin x and what is the answer to the inequality?**
To solve the inequality \(\sin(2x) \geq \sin x\), follow these steps:
1. **Use Trigonometric Identities:**
- Remember the identity for \(\sin(2x)\):
\[
\sin(2x) = 2\sin x \cos x
\]
- Substitute it into the inequality:
\[
2\sin x \cos x \geq \sin x
\]
2. **Simplify the Inequality:**
- Factor out \(\sin x\) from both sides:
\[
\sin x (2\cos x - 1) \geq 0
\]
3. **Analyze Critical Points:**
- Consider the cases where the product \(\sin x (2\cos x - 1)\) is zero:
- \(\sin x = 0\): This gives \(x = n\pi\) where \(n\) is an integer.
- \(2\cos x - 1 = 0\): Solving gives \(\cos x = \frac{1}{2}\), which implies \(x = \frac{\pi}{3} + 2k\pi\) or \(x = -\frac{\pi}{3} + 2k\pi\) for any integer \(k\).
4. **Determine Intervals:**
- Test intervals determined by critical points:
- When \(\sin x > 0\) and \(2\cos x - 1 \geq 0\), the inequality holds.
- When \(\sin x < 0\) and \(2\cos x - 1 \leq 0\), the inequality also holds.
5. **Final Solution:**
- Combine all the conditions to find the solution set.
The solution indicates where the inequality is held true across different intervals of \(x\). Analyze specific intervals to derive exact solutions for each given range using knowledge of the unit circle and periodicity of sine and cosine functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcaf0d7ce-11be-486e-aed4-80125ae9c37f%2Fd9565d7f-c9f3-4ba4-84b4-93946efe9832%2F0rgz6g_processed.png&w=3840&q=75)
Transcribed Image Text:**How would you solve the inequality sin(2x) ≥ sin x and what is the answer to the inequality?**
To solve the inequality \(\sin(2x) \geq \sin x\), follow these steps:
1. **Use Trigonometric Identities:**
- Remember the identity for \(\sin(2x)\):
\[
\sin(2x) = 2\sin x \cos x
\]
- Substitute it into the inequality:
\[
2\sin x \cos x \geq \sin x
\]
2. **Simplify the Inequality:**
- Factor out \(\sin x\) from both sides:
\[
\sin x (2\cos x - 1) \geq 0
\]
3. **Analyze Critical Points:**
- Consider the cases where the product \(\sin x (2\cos x - 1)\) is zero:
- \(\sin x = 0\): This gives \(x = n\pi\) where \(n\) is an integer.
- \(2\cos x - 1 = 0\): Solving gives \(\cos x = \frac{1}{2}\), which implies \(x = \frac{\pi}{3} + 2k\pi\) or \(x = -\frac{\pi}{3} + 2k\pi\) for any integer \(k\).
4. **Determine Intervals:**
- Test intervals determined by critical points:
- When \(\sin x > 0\) and \(2\cos x - 1 \geq 0\), the inequality holds.
- When \(\sin x < 0\) and \(2\cos x - 1 \leq 0\), the inequality also holds.
5. **Final Solution:**
- Combine all the conditions to find the solution set.
The solution indicates where the inequality is held true across different intervals of \(x\). Analyze specific intervals to derive exact solutions for each given range using knowledge of the unit circle and periodicity of sine and cosine functions.
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