Задание №86922 ЕГЭ Профильной математике

$$\begin{gathered} \text{а)} {\sin }^2\left(\frac{\mathrm{х}}{2}-\frac{\mathrm{\pi }}{4}\right) = \left(\right(\sin \frac{\mathrm{х}}{2}\left)\cos \right(\frac{\mathrm{\pi }}{4}) - \cos (\frac{\mathrm{х}}{2})\sin {\left(\frac{\mathrm{\pi }}{4}\right))}^2= \\ ={(\frac{1}{\sqrt{2}}\sin \frac{\mathrm{х}}{2} - \frac{1}{\sqrt{2}} \cos \frac{\mathrm{х}}{2})}^2 =\frac{1}{2}{\left(\sin \left(\frac{\mathrm{х}}{2}\right)-\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2 {\sin }^2\left(\frac{\mathrm{х}}{2}+\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}{\left(\sin \left(\frac{\mathrm{х}}{2}\right)+\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2 \\ 2·\frac{1}{2}{\left(\sin \left(\frac{\mathrm{х}}{2}\right)-\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2·\frac{1}{2}{\left(\sin \left(\frac{\mathrm{х}}{2}\right)+\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2=\frac{1}{2}{\left(\sin \left(\frac{\mathrm{х}}{2}\right)-\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2·{\left(\sin \left(\frac{\mathrm{х}}{2}\right)+\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2= \\ =\frac{1}{2}{\left(\sin \left(\frac{\mathrm{х}}{2}\right)-\cos \left(\frac{\mathrm{х}}{2}\right)·\sin \left(\frac{\mathrm{х}}{2}\right)+\cos \left(\frac{\mathrm{х}}{2}\right)\right)}^2=\frac{1}{2}{\left({\sin }^2\left(\frac{\mathrm{х}}{2}\right)-{\cos }^2\left(\frac{\mathrm{х}}{2}\right)\right)}^2=\frac{1}{2}\left(-\cos \left(2·\frac{\mathrm{х}}{2}\right)\right)=\frac{1}{2}{\cos }^2\left(\mathrm{х}\right). \end{gathered}$$

$$\begin{gathered} \frac{1}{2}{\cos }^2\left(\mathrm{х}\right)={\cos }^4\left(\mathrm{х}\right) \\ \frac{1}{2}{\cos }^2\left(\mathrm{х}\right)-{\cos }^4\left(\mathrm{х}\right)=0 \\ {\cos }^2\left(\mathrm{х}\right)(\frac{1}{2}-{\cos }^2\left(\mathrm{х}\right))=0 \\ {\cos }^2\left(\mathrm{х}\right)=0 \frac{1}{2}-{\cos }^2\left(\mathrm{х}\right)=0 \\ \cos \left(\mathrm{х}\right)=0 {\cos }^2\left(\mathrm{х}\right)=\frac{1}{2} \\ {\mathrm{х}}_1=\frac{\mathrm{\pi }}{2}+\mathrm{\pi k},\mathrm{k}\in \mathrm{Z} \cos \left(\mathrm{х}\right)=\pm \frac{1}{\sqrt{2}} \\ {\mathrm{x}}_2=\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi n}}{2}, \mathrm{n}\in \mathrm{Z} \end{gathered}$$

$\text{б)} [-3\mathrm{\pi }; -2\mathrm{\pi }]$

$$\begin{gathered} {\mathrm{х}}_3=-3\mathrm{\pi }+\frac{\mathrm{\pi }}{4}=\frac{-11\mathrm{\pi }}{4} \\ {\mathrm{х}}_4=-3\mathrm{\pi }+\frac{\mathrm{\pi }}{2}=\frac{-5\mathrm{\pi }}{2} \\ {\mathrm{х}}_5=-2\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{-9\mathrm{\pi }}{4} \end{gathered}$$