29. $$ \sec ^{2} x-\sec x=2 $$ 30. $$ \sec x \csc x=2 \csc x $$ 31. $$ 2 \sin x+\csc x=0 $$

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Certainly! Let's solve each of the given trigonometric equations step by step.

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## 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:
   $$
   x = \frac{\pi}{3} + 2\pi k \quad \text{and} \quad x = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is an integer}
   $$

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## 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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If you have any further questions or need additional clarification on these problems, feel free to ask!
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.

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