Answer
Solutions:
1. **Problem 29:**
\[
x = \frac{\pi}{3} + 2\pi k, \quad \frac{5\pi}{3} + 2\pi k, \quad \text{or} \quad \pi + 2\pi k \quad \text{where } k \text{ is an integer}
\]
2. **Problem 30:**
\[
x = \frac{\pi}{3} + 2\pi k \quad \text{and} \quad x = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is an integer}
\]
3. **Problem 31:**
No real solutions exist.
Solution
Certainly! Let's solve each of the given trigonometric equations step by step.
---
## Problem 29
**Equation:**
\[
\sec^{2} x - \sec x = 2
\]
**Solution:**
1. **Substitute \( \sec x \) with a variable:**
Let \( y = \sec x \). The equation becomes:
\[
y^2 - y = 2
\]
2. **Rearrange the equation:**
\[
y^2 - y - 2 = 0
\]
3. **Solve the quadratic equation:**
\[
y = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}
\]
So, \( y = 2 \) or \( y = -1 \).
4. **Find \( x \) for each case:**
- **Case 1:** \( \sec x = 2 \)
\[
\cos x = \frac{1}{2}
\]
The solutions for \( \cos x = \frac{1}{2} \) are:
\[
x = \frac{\pi}{3} + 2\pi k \quad \text{and} \quad x = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is an integer}
\]
- **Case 2:** \( \sec x = -1 \)
\[
\cos x = -1
\]
The solution for \( \cos x = -1 \) is:
\[
x = \pi + 2\pi k \quad \text{where } k \text{ is an integer}
\]
5. **General Solution:**
\[
x = \frac{\pi}{3} + 2\pi k, \quad \frac{5\pi}{3} + 2\pi k, \quad \text{or} \quad \pi + 2\pi k \quad \text{where } k \text{ is an integer}
\]
---
## Problem 30
**Equation:**
\[
\sec x \csc x = 2 \csc x
\]
**Solution:**
1. **Assume \( \csc x \neq 0 \):**
Since \( \csc x = \frac{1}{\sin x} \), \( \sin x \neq 0 \).
2. **Divide both sides by \( \csc x \):**
\[
\sec x = 2
\]
3. **Solve for \( x \):**
\[
\sec x = 2 \implies \cos x = \frac{1}{2}
\]
4. **Find \( x \):**
The solutions for \( \cos x = \frac{1}{2} \) are:
\[
x = \frac{\pi}{3} + 2\pi k \quad \text{and} \quad x = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is an integer}
\]
---
## Problem 31
**Equation:**
\[
2 \sin x + \csc x = 0
\]
**Solution:**
1. **Express \( \csc x \) in terms of \( \sin x \):**
\[
2 \sin x + \frac{1}{\sin x} = 0
\]
2. **Multiply both sides by \( \sin x \) (assuming \( \sin x \neq 0 \)):**
\[
2 \sin^2 x + 1 = 0
\]
3. **Rearrange the equation:**
\[
2 \sin^2 x = -1 \implies \sin^2 x = -\frac{1}{2}
\]
4. **Analyze the equation:**
\( \sin^2 x \) is always non-negative for real numbers, but the equation \( \sin^2 x = -\frac{1}{2} \) implies that \( \sin^2 x \) is negative, which is impossible.
5. **Conclusion:**
There are **no real solutions** to the equation \( 2 \sin x + \csc x = 0 \).
---
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