$ \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta $[NEWLINE] $\cos\left(x + \dfrac{\pi}{4}\right) = $$\cos x \cdot \dfrac{\sqrt{2}}{2} - \sin x \cdot \dfrac{\sqrt{2}}{2} = $$\dfrac{\sqrt{2}}{2}(\cos x - \sin x)$[NEWLINE] $ \left(\cos\left(x + \dfrac{\pi}{4}\right)\right)^2 = $$\left(\dfrac{\sqrt{2}}{2}\right)^2 (\cos x - \sin x)^2 = $$\dfrac{1}{2}(\cos^2 x - 2\cos x \sin x + \sin^2 x) = $$\dfrac{1}{2}(1 - 2\cos x \sin x) $[NEWLINE] \(\)[NEWLINE] Тогда:[NEWLINE] $ \sin^2 x + \dfrac{1}{2}(1 - 2\cos x \sin x) = \dfrac{1}{2} $[NEWLINE] $ \sin^2 x + \dfrac{1}{2} - \cos x \sin x = \dfrac{1}{2} $[NEWLINE] $ \sin^2 x - \cos x \sin x = 0 $[NEWLINE] $ \sin x (\sin x - \cos x) = 0 $[NEWLINE] [TWO-COLUMNS]$\sin x = 0$[NEWLINE]$x = \pi n, \quad n \in Z$[COLUMN-BREAK]$ \sin x - \cos x=0|:\cos x$[NEWLINE]$tgx=1$[NEWLINE]$ x = \dfrac{\pi}{4} + \pi k, \quad k \in Z$[/TWO-COLUMNS] [image:0:right]б) Отберем корни, принадлежащие отрезку $\left[\dfrac{13\pi}{2};\; \dfrac{15\pi}{2}\right].$[IMAGENEWLINE] $x_1 = 7\pi$[IMAGENEWLINE] $x_2 = 7\pi + \frac{\pi}{4} = \frac{29\pi}{4}$[IMAGENEWLINE][IMAGENEWLINE] [BOLD]Ответ:[/BOLD] а) $x = \pi n, \ n \in \mathbb{Z}, \ x = \pm \frac{\pi}{4} + \pi k, \ k \in \mathbb{Z}$[IMAGENEWLINE] б) $7\pi, \frac{29\pi}{4}$[IMAGENEWLINE]