The problem you’ve uploaded is:

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 $1$).