29. 4 cos 4x = √2/2 , [0; π].

$$cos(4x) = \frac{\sqrt{2}}{2}$$

$$4x = \pm \frac{\pi}{4} + 2\pi n, n \in Z$$

$$x = \pm \frac{\pi}{16} + \frac{\pi n}{2}, n \in Z$$

Найдем корни уравнения на заданном отрезке $$[0; \pi]$$

$$x = \frac{\pi}{16} + \frac{\pi \cdot 0}{2} = \frac{\pi}{16}$$

$$x = \frac{\pi}{16} + \frac{\pi \cdot 1}{2} = \frac{\pi}{16} + \frac{8\pi}{16} = \frac{9\pi}{16}$$

$$x = \frac{\pi}{16} + \frac{\pi \cdot 2}{2} = \frac{\pi}{16} + \frac{16\pi}{16} = \frac{17\pi}{16} > \pi$$

$$x = -\frac{\pi}{16} + \frac{\pi \cdot 1}{2} = -\frac{\pi}{16} + \frac{8\pi}{16} = \frac{7\pi}{16}$$

$$x = -\frac{\pi}{16} + \frac{\pi \cdot 2}{2} = -\frac{\pi}{16} + \frac{16\pi}{16} = \frac{15\pi}{16}$$

Ответ: $$x = \frac{\pi}{16}, x = \frac{7\pi}{16}, x = \frac{9\pi}{16}, x = \frac{15\pi}{16}$$