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**Solve, for \( 0 \leq \theta \leq 360^\circ \), the equation:**
\[
\tan^2\theta + \sec^2\theta + 5\sec\theta = 2
\]
**Give your answers to 1 decimal place.**

### Steps to solve the equation:

1. **Substitute trigonometric identity:**
   Recall that \(\tan^2\theta + 1 = \sec^2\theta\). Thus, we can replace \(\tan^2\theta + \sec^2\theta\) in the equation:
   \[
   \tan^2\theta + \sec^2\theta = \sec^2\theta - 1 + \sec^2\theta = 2\sec^2\theta - 1
   \]

   Now the equation becomes:
   \[
   2\sec^2\theta - 1 + 5\sec\theta = 2
   \]

2. **Simplify:**
   Move constants to the right-hand side:
   \[
   2\sec^2\theta + 5\sec\theta - 3 = 0
   \]

3. **Let \( x = \sec\theta \):**
   This transforms the equation into a quadratic equation:
   \[
   2x^2 + 5x - 3 = 0
   \]

4. **Solve the quadratic equation:**
   Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 5\), and \(c = -3\):
   \[
   x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}
   \]
   \[
   x = \frac{-5 \pm \sqrt{25 + 24}}{4}
   \]
   \[
   x = \frac{-5 \pm \sqrt{49}}{4}
   \]
   \[
   x = \frac{-5 \pm 7}{4}
   \]
   Thus, \(x = \frac{-5 + 7}{4} = \frac{2}{4} = 0.5\), or \(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\).

   Therefore, \( \sec\theta = 0.5 \) or \( \sec\theta = -3 \).

5. **Find the corresponding angles:**
   Recall that \(\sec\theta = \frac{1}{\cos\theta}\), so:
   
   - If \( \sec\theta = 0.5 \), then \( \cos\theta = 2 \), which is impossible since the cosine function is always between \(-1\) and \(1\).
   - If \( \sec\theta = -3 \), then \( \cos\theta = -\frac{1}{3} \).

   For \( \cos\theta = -\frac{1}{3} \), find the angles:
   \[
   \theta = \cos^{-1}\left(-\frac{1}{3}\right)
   \]

   Using a calculator:
   \[
   \theta \approx 109.5^\circ \quad \text{or} \quad \theta \approx 250.5^\circ
   \]

### Final answer:
The solutions for \( \theta \), correct to one decimal place, are:
\[
\theta = 109.5^\circ, \, 250.5^\circ
\]

Would you like more details or have any questions?

Here are 5 related questions for further practice:
1. What is the general solution to the equation \( \sec\theta = -3 \)?
2. How would you solve the equation if \(\sec^2 \theta + 4 \sec\theta - 5 = 0\)?
3. Can you solve \( \cos 2\theta + \sin \theta = 0 \) for \(0^\circ \leq \theta \leq 360^\circ\)?
4. How do you convert trigonometric equations to quadratic form?
5. What are the key differences between solving trigonometric and polynomial equations?

**Tip:** Always check if the solutions fall within the allowed range for trigonometric functions (e.g., \(\cos \theta\) values must be between \(-1\) and \(1\)).

Solve, for 0 ≤ θ ≤ 360, the equation: tan^2 θ + sec^2 θ + 5 sec θ = 2

Explore the solution to the trigonometric equation tan^2 θ + sec^2 θ + 5 sec θ = 2 for 0° ≤ θ ≤ 360°. This problem involves trigonometric identities and solving quadratic equations, providing a detailed step-by-step solution. Ideal for advanced students in Grades 11-12.

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