$ \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta $[NEWLINE] $\sin\left(x - \dfrac{\pi}{4}\right) = $$\sin x \cdot \dfrac{\sqrt{2}}{2} - \cos x \cdot \dfrac{\sqrt{2}}{2} = $$\dfrac{\sqrt{2}}{2}(\sin x - \cos x)$[NEWLINE] $ \left(\sin\left(x - \dfrac{\pi}{4}\right)\right)^2 = $$\left(\dfrac{\sqrt{2}}{2}\right)^2 (\sin x - \cos x)^2 = $$\dfrac{1}{2}(\sin^2 x - 2\sin x \cos x + \cos^2 x) = $$\dfrac{1}{2}(1 - 2\sin x \cos x) $[NEWLINE] \(\)[NEWLINE] Тогда:[NEWLINE] $ \cos^2 x + \dfrac{1}{2}(1 - 2\sin x \cos x) = \dfrac{1}{2} $[NEWLINE] $ \cos^2 x + \dfrac{1}{2} - \sin x \cos x = \dfrac{1}{2} $[NEWLINE] $ \cos^2 x - \sin x \cos x = 0 $[NEWLINE] $ \cos x (\cos x - \sin x) = 0 $[NEWLINE] [TWO-COLUMNS]$\cos x = 0$[NEWLINE]$x = \dfrac{\pi}{2} + \pi n, \quad n \in Z$[COLUMN-BREAK]$ \cos x - \sin 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 $\left[5\pi;\; 6\pi\right].$[IMAGENEWLINE] $5\pi = \dfrac{10\pi}{2}, \quad 6\pi = \dfrac{12\pi}{2}$[IMAGENEWLINE] $x_1 = \dfrac{11\pi}{2}$[IMAGENEWLINE] $x_2 = 5\pi + \frac{\pi}{4} = \frac{21\pi}{4}$[IMAGENEWLINE][IMAGENEWLINE] [BOLD]Ответ:[/BOLD] а) $x = \dfrac{\pi}{2} + \pi n, \ n \in \mathbb{Z}, \ x = \dfrac{\pi}{4} + \pi k, \ k \in \mathbb{Z}$[IMAGENEWLINE] б) $\dfrac{11\pi}{2}, \dfrac{21\pi}{4}$[IMAGENEWLINE]