Simplify and solve the equation: \( \tan(4x) = 1 \) for all solutions of \( x \) in the range \( (0, 2\pi) \).

To solve the equation \( \tan(4x) = 1 \), we first recognize that the tangent function equals 1 at specific angles. Specifically, we have: \[ 4x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \] Next, we can solve for \( x \): \[ x = \frac{\pi}{16} + \frac{n\pi}{4} \] Now, we need to find values of \( n \) such that \( x \) is in the range \( (0, 2\pi) \). Calculating the possible values of \( x \): 1. For \( n = 0 \): \[ x = \frac{\pi}{16} \quad (\text{which is in } (0, 2\pi)) \] 2. For \( n = 1 \): \[ x = \frac{\pi}{16} + \frac{\pi}{4} = \frac{\pi}{16} + \frac{4\pi}{16} = \frac{5\pi}{16} \quad (\text{which is in } (0, 2\pi)) \] 3. For \( n = 2 \): \[ x = \frac{\pi}{16} + \frac{2\pi}{4} = \frac{\pi}{16} + \frac{8\pi}{16} = \frac{9\pi}{16} \quad (\text{which is in } (0, 2\pi)) \] 4. For \( n = 3 \): \[ x = \frac{\pi}{16} + \frac{3\pi}{4} = \frac{\pi}{16} + \frac{12\pi}{16} = \frac{13\pi}{16} \quad (\text{which is in } (0, 2\pi)) \] 5. For \( n = 4 \): \[ x = \frac{\pi}{16} + \frac{4\pi}{4} = \frac{\pi}{16} + \frac{16\pi}{16} = \frac{17\pi}{16} \quad (\text{which is in } (0, 2\pi)) \] 6. For \( n = 5 \): \[ x = \frac{\pi}{16} + \frac{5\pi}{4} = \frac{\pi}{16} + \frac{20\pi}{16} = \frac{21\pi}{16} \quad (\text{which is in } (0, 2\pi)) \] 7. For \( n = 6 \): \[ x = \frac{\pi}{16} + \frac{6\pi}{4} = \frac{\pi}{16} + \frac{24\pi}{16} = \frac{25\pi}{16} \quad (\text{which is not in } (0, 2\pi)) \] Thus, the valid values of \( x \) in the range \( (0, 2\pi) \) are: \[ x = \frac{\pi}{16}, \frac{5\pi}{16}, \frac{9\pi}{16}, \frac{13\pi}{16}, \frac{17\pi}{16}, \frac{21\pi}{16}. \] In summary, the complete solution set is: \[ x = \frac{\pi}{16}, \frac{5\pi}{16}, \frac{9\pi}{16}, \frac{13\pi}{16}, \frac{17\pi}{16}, \frac{21\pi}{16}. \]