Let's go through the given problems.

The text asks to find the exact solutions of the trigonometric equations within the interval $[0, 2\pi]$.

Problem 2:

$\cos^2\left(2x - \frac{\pi}{3}\right) = 1$

We start by simplifying this equation. The cosine squared of an angle equals 1 when $\cos$ of that angle equals either 1 or -1.

1. $\cos\left(2x - \frac{\pi}{3}\right) = \pm 1$

This gives us two cases to solve:

• Case 1: $\cos\left(2x - \frac{\pi}{3}\right) = 1$
• Case 2: $\cos\left(2x - \frac{\pi}{3}\right) = -1$

Case 1: $\cos\left(2x - \frac{\pi}{3}\right) = 1$

The general solution for $\cos(\theta) = 1$ is: $2x - \frac{\pi}{3} = 2k\pi, \text{ where } k \in \mathbb{Z}$ Solving for $x$: $2x = 2k\pi + \frac{\pi}{3}$ $x = k\pi + \frac{\pi}{6}$ Now we check which values of $x$ are within $[0, 2\pi]$:

• For $k = 0$, $x = \frac{\pi}{6}$.
• For $k = 1$, $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
• For $k = 2$, $x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$, which is greater than $2\pi$.

Thus, the solutions for Case 1 are: $x = \frac{\pi}{6}, \frac{7\pi}{6}$

Case 2: $\cos\left(2x - \frac{\pi}{3}\right) = -1$

The general solution for $\cos(\theta) = -1$ is: $2x - \frac{\pi}{3} = (2k+1)\pi, \text{ where } k \in \mathbb{Z}$ Solving for $x$: $2x = (2k+1)\pi + \frac{\pi}{3}$ $x = \frac{(2k+1)\pi}{2} + \frac{\pi}{6}$ Now, check which values of $x$ are within $[0, 2\pi]$:

• For $k = 0$, $x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3}$.
• For $k = 1$, $x = \frac{3\pi}{2} + \frac{\pi}{6} = \frac{11\pi}{6}$.

Thus, the solutions for Case 2 are: $x = \frac{2\pi}{3}, \frac{11\pi}{6}$

Final solution for Problem 2:

$x = \frac{\pi}{6}, \frac{7\pi}{6}, \frac{2\pi}{3}, \frac{11\pi}{6}$

Problem 3:

$\cos\left(2x - \frac{\pi}{5}\right) = -1$

The general solution for $\cos(\theta) = -1$ is: $2x - \frac{\pi}{5} = (2k+1)\pi, \text{ where } k \in \mathbb{Z}$ Solving for $x$: $2x = (2k+1)\pi + \frac{\pi}{5}$ $x = \frac{(2k+1)\pi}{2} + \frac{\pi}{10}$ Now, check which values of $x$ are within $[0, 2\pi]$:

• For $k = 0$, $x = \frac{\pi}{2} + \frac{\pi}{10} = \frac{3\pi}{5}$.
• For $k = 1$, $x = \frac{3\pi}{2} + \frac{\pi}{10} = \frac{17\pi}{10} = 1.7\pi$, which is still less than $2\pi$.

Thus, the solutions for Problem 3 are: $x = \frac{3\pi}{5}, \frac{17\pi}{10}$

Let me know if you'd like further clarifications!

Five Related Questions:

1. How do you determine the general solutions for trigonometric equations?
2. Why do we consider the interval $[0, 2\pi]$ when solving trigonometric problems?
3. What is the difference between the general solutions for $\cos(\theta) = 1$ and $\cos(\theta) = -1$?
4. How do transformations inside the cosine function, such as $2x - \frac{\pi}{3}$, affect the solution set?
5. Can these methods be used for solving $\sin$ and $\tan$ equations?

Tip:

When solving trigonometric equations, always check the principal range (like $[0, 2\pi]$) to avoid extra or missed solutions.