6) cos (\frac{π}{4}+x)+cos(\frac{π}{4}−x)=1;

Используем формулу суммы косинусов: $$cos a + cos b = 2 cos \frac{a + b}{2} cos \frac{a - b}{2}$$.

В нашем случае $$a = \frac{\pi}{4} + x, b = \frac{\pi}{4} - x$$.

$$cos (\frac{\pi}{4}+x)+cos(\frac{\pi}{4}−x) = 2 cos \frac{(\frac{\pi}{4}+x) + (\frac{\pi}{4} - x)}{2} cos \frac{(\frac{\pi}{4}+x) - (\frac{\pi}{4} - x)}{2} = 1$$

$$2 cos \frac{\frac{\pi}{2}}{2} cos \frac{2x}{2} = 1$$

$$2 cos \frac{\pi}{4} cos x = 1$$

$$2 \cdot \frac{\sqrt{2}}{2} cos x = 1$$

$$\sqrt{2} cos x = 1$$

$$cos x = \frac{1}{\sqrt{2}}$$

$$cos x = \frac{\sqrt{2}}{2}$$

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

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