To solve the equation $2\sec(x) = \tan(x) + \cot(x)$ for $x$ in the interval $0 \leq x \leq 2\pi$, let's proceed step by step.

Step 1: Simplify the equation

First, recall the following trigonometric identities: $\sec(x) = \frac{1}{\cos(x)}, \quad \tan(x) = \frac{\sin(x)}{\cos(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)}$

Substituting these into the equation: $2\cdot\frac{1}{\cos(x)} = \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}$

Step 2: Multiply through by $\cos(x)$

To eliminate the denominator $\cos(x)$, multiply every term by $\cos(x)$: $2 = \sin(x) + \frac{\cos^2(x)}{\sin(x)}$

Step 3: Multiply through by $\sin(x)$

To simplify further, multiply the entire equation by $\sin(x)$: $2\sin(x) = \sin^2(x) + \cos^2(x)$

Using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, the equation becomes: $2\sin(x) = 1$

Step 4: Solve for $\sin(x)$

$\sin(x) = \frac{1}{2}$

Step 5: Find the values of $x$

The general solution for $\sin(x) = \frac{1}{2}$ in the interval $0 \leq x \leq 2\pi$ occurs at: $x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}$

Final Answer:

The solutions to the equation $2\sec(x) = \tan(x) + \cot(x)$ within the interval $0 \leq x \leq 2\pi$ are: $x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}$

Would you like more details on any of these steps, or have any other questions?

Here are some related questions you might explore:

1. What is the general solution for $\sin(x) = \frac{1}{2}$?
2. How can you derive the identity $\sin^2(x) + \cos^2(x) = 1$?
3. What is the domain and range of the secant function?
4. How do you solve trigonometric equations involving multiple angles?
5. What is the significance of the interval $0 \leq x \leq 2\pi$ in trigonometric equations?

Tip: Always remember to check if the solutions fall within the given interval when solving trigonometric equations.