To solve the equation $\sec(\theta) = \sqrt{2}$ on the interval $0 \leq \theta < 2\pi$, we'll follow these steps:

1. Understanding the Secant Function

Recall that the secant function is the reciprocal of the cosine function:

$\sec(\theta) = \frac{1}{\cos(\theta)}$

So, $\sec(\theta) = \sqrt{2}$ implies:

$\cos(\theta) = \frac{1}{\sqrt{2}}$

Now, we will solve for $\theta$ by identifying the angles where $\cos(\theta) = \frac{1}{\sqrt{2}}$.

2. Finding the Angles

The value $\cos(\theta) = \frac{1}{\sqrt{2}}$ corresponds to two specific angles within one full rotation, $0 \leq \theta < 2\pi$:

• $\theta = \frac{\pi}{4}$ (or 45°)
• $\theta = \frac{7\pi}{4}$ (or 315°)

Thus, the solutions to the equation $\sec(\theta) = \sqrt{2}$ within the interval $0 \leq \theta < 2\pi$ are:

$\theta = \frac{\pi}{4}, \frac{7\pi}{4}$

3. Answer in Degrees and Radians

• In radians, the solutions are: $\frac{\pi}{4}$ and $\frac{7\pi}{4}$
• In degrees, the solutions are: 45° and 315°

4. Type of Line

To represent this on a unit circle, you would draw two lines:

• One at $\theta = \frac{\pi}{4}$ (or 45°), which is a diagonal line in the first quadrant.
• One at $\theta = \frac{7\pi}{4}$ (or 315°), which is another diagonal line in the fourth quadrant.

These are diagonal lines since $\cos(\theta)$ corresponds to the horizontal distance from the origin on the unit circle.

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Let me know if you'd like further details or if you have any questions! Here's some follow-up you might find interesting:

1. What is the relationship between secant and cosine, and how does this help solve the equation?
2. Can you explain why there are two solutions within one full rotation?
3. How would the solutions change if the equation were $\sec(\theta) = -\sqrt{2}$?
4. Can secant be negative, and if so, where would those solutions lie on the unit circle?
5. How do you graph secant and cosine functions?

Tip: Always remember that the secant function is positive when the cosine function is positive and negative when the cosine function is negative. This will help you identify solution intervals on the unit circle.