![cos2x-1=\sqrt2sin(\frac{5\pi}{2}-x)\\-2sin^2x=\sqrt2sin(2\pi+(\frac{\pi}{2}-x))\\-2(1-cos^2x)=\sqrt2sin(\frac{\pi}{2}-x)\\-2+2cos^2x=\sqrt2cosx\\cosx=u\\2u^2-\sqrt2u-2=0\\D:2+16=18\\x_1,_2=\frac{\sqrt2\pm3\sqrt2}{4}\\\\x_1=\sqrt2\\cosx \in[-1;1], \quad cosx \neq \sqrt2;\\\\x_2=-\frac{\sqrt2}{2}\\cosx=-\frac{\sqrt2}{2}\\x=\pm(\pi-arccos\frac{\sqrt2}{2})+2\pi n\\x=\pm(\pi-\frac{\pi}{4})+2\pi n\\x=\pm\frac{3\pi}{4}+2\pi n, \; n\in Z;](https://tex.z-dn.net/?f=cos2x-1%3D%5Csqrt2sin%28%5Cfrac%7B5%5Cpi%7D%7B2%7D-x%29%5C%5C-2sin%5E2x%3D%5Csqrt2sin%282%5Cpi%2B%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%29%29%5C%5C-2%281-cos%5E2x%29%3D%5Csqrt2sin%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%29%5C%5C-2%2B2cos%5E2x%3D%5Csqrt2cosx%5C%5Ccosx%3Du%5C%5C2u%5E2-%5Csqrt2u-2%3D0%5C%5CD%3A2%2B16%3D18%5C%5Cx_1%2C_2%3D%5Cfrac%7B%5Csqrt2%5Cpm3%5Csqrt2%7D%7B4%7D%5C%5C%5C%5Cx_1%3D%5Csqrt2%5C%5Ccosx+%5Cin%5B-1%3B1%5D%2C+%5Cquad+cosx+%5Cneq+%5Csqrt2%3B%5C%5C%5C%5Cx_2%3D-%5Cfrac%7B%5Csqrt2%7D%7B2%7D%5C%5Ccosx%3D-%5Cfrac%7B%5Csqrt2%7D%7B2%7D%5C%5Cx%3D%5Cpm%28%5Cpi-arccos%5Cfrac%7B%5Csqrt2%7D%7B2%7D%29%2B2%5Cpi+n%5C%5Cx%3D%5Cpm%28%5Cpi-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2B2%5Cpi+n%5C%5Cx%3D%5Cpm%5Cfrac%7B3%5Cpi%7D%7B4%7D%2B2%5Cpi+n%2C+%5C%3B+n%5Cin+Z%3B "cos2x-1=\sqrt2sin(\frac{5\pi}{2}-x)\\-2sin^2x=\sqrt2sin(2\pi+(\frac{\pi}{2}-x))\\-2(1-cos^2x)=\sqrt2sin(\frac{\pi}{2}-x)\\-2+2cos^2x=\sqrt2cosx\\cosx=u\\2u^2-\sqrt2u-2=0\\D:2+16=18\\x_1,_2=\frac{\sqrt2\pm3\sqrt2}{4}\\\\x_1=\sqrt2\\cosx \in[-1;1], \quad cosx \neq \sqrt2;\\\\x_2=-\frac{\sqrt2}{2}\\cosx=-\frac{\sqrt2}{2}\\x=\pm(\pi-arccos\frac{\sqrt2}{2})+2\pi n\\x=\pm(\pi-\frac{\pi}{4})+2\pi n\\x=\pm\frac{3\pi}{4}+2\pi n, \; n\in Z;")
Можно найти корни принадлежащие заданному отрезку подставляя целые числа за n и вычисления покажут какие из корней принадлежат отрезку, а какие нет. Это будет выглядеть так:
Или другой способ, что проще, найти корни на координатной прямой: снимок во вложении.
