2*х2-квадрат -2*у2-квадрат :(х-у)(х+у)=2х2-кв. -2у2-кв. :2х-ху-ух-у2-кв.
Если х=2;у=3, то 2*2 2-кв. - 2*3 2-кв. :2*2-2*3-3*2-3 2-кв. =2*4-2*9:4-6-6-9=8-18:4-6-6-9=8-4,5-12-9=3,5-12-9=-8,5-9=-18
Ответ: -18
Косинус отрицателен во 2 и 3 четверти, значит
![\cos( \pi + \frac{\pi}{4} )=\cos \frac{5 \pi }{4} =- \frac{ \sqrt{2} } {2} \\ \\ \cos(\pi -\frac{\pi}{4} )= \cos\frac{3 \pi }{4} =- \frac{ \sqrt{2} }{2} \\ \\ \cos(3 \pi -\frac{\pi}{4} )=\cos\frac{11\pi}{4} =- \frac{ \sqrt{2} }{2}](https://tex.z-dn.net/?f=%5Ccos%28+%5Cpi+%2B+%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%3D%5Ccos+%5Cfrac%7B5+%5Cpi+%7D%7B4%7D+%3D-+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D+%7B2%7D+%5C%5C+%5C%5C+%5Ccos%28%5Cpi+-%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%3D+%5Ccos%5Cfrac%7B3+%5Cpi+%7D%7B4%7D+%3D-+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D+%5C%5C+%5C%5C+%5Ccos%283+%5Cpi+-%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%3D%5Ccos%5Cfrac%7B11%5Cpi%7D%7B4%7D+%3D-+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D+)
1/2≤cos(2x-π/3)≤1
-π/3+2πk≤2x-π/3≤π/3+2πk
2πk≤2x≤2π/3+2πk
πk≤x≤π/3+πk