1) а)2+5а+3а-9=8а-7
б)y^2-4y+5-y^2+5y-8= -y-3= -(y+3)
в)3х+4х^2-2x^2+6x= 2x^2+9x
![4cos^2(\frac{5\pi}{4}-x)=1\\\\cos^2(\frac{5\pi}{4}-x)=\frac{1}{4} \\\\cos^2 \alpha = \frac{1+cos2 \alpha }{2}\; \; \Rightarrow \; \; cos^2(\frac{5\pi}{4}-x)= \frac{1+cos(\frac{5\pi}{2}-2x)}{2} =\frac{1}{4} \\\\1+cos(\frac{5\pi}{2}-2x)=\frac{1}{2}](https://tex.z-dn.net/?f=4cos%5E2%28%5Cfrac%7B5%5Cpi%7D%7B4%7D-x%29%3D1%5C%5C%5C%5Ccos%5E2%28%5Cfrac%7B5%5Cpi%7D%7B4%7D-x%29%3D%5Cfrac%7B1%7D%7B4%7D+%5C%5C%5C%5Ccos%5E2+%5Calpha+%3D+%5Cfrac%7B1%2Bcos2+%5Calpha+%7D%7B2%7D%5C%3B+%5C%3B+%5CRightarrow+%5C%3B+%5C%3B+cos%5E2%28%5Cfrac%7B5%5Cpi%7D%7B4%7D-x%29%3D+%5Cfrac%7B1%2Bcos%28%5Cfrac%7B5%5Cpi%7D%7B2%7D-2x%29%7D%7B2%7D+%3D%5Cfrac%7B1%7D%7B4%7D+%5C%5C%5C%5C1%2Bcos%28%5Cfrac%7B5%5Cpi%7D%7B2%7D-2x%29%3D%5Cfrac%7B1%7D%7B2%7D)
![cos(\frac{5\pi}{2}-2x)=-\frac{1}{2}\; \; \Rightarrow \; \; sin2x=-\frac{1}{2}\\\\2x=(-1)^{k}arcsin(-\frac{1}{2})+\pi k=(-1)^{k+1}\cdot \frac{\pi}{6}+\pi k,\; k\in Z\\\\x=(-1)^{k+1}\cdot \frac{\pi}{12}+\frac{\pi k}{2}\; ,k\in Z](https://tex.z-dn.net/?f=cos%28%5Cfrac%7B5%5Cpi%7D%7B2%7D-2x%29%3D-%5Cfrac%7B1%7D%7B2%7D%5C%3B+%5C%3B+%5CRightarrow+%5C%3B+%5C%3B+sin2x%3D-%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C2x%3D%28-1%29%5E%7Bk%7Darcsin%28-%5Cfrac%7B1%7D%7B2%7D%29%2B%5Cpi+k%3D%28-1%29%5E%7Bk%2B1%7D%5Ccdot+%5Cfrac%7B%5Cpi%7D%7B6%7D%2B%5Cpi+k%2C%5C%3B+k%5Cin+Z%5C%5C%5C%5Cx%3D%28-1%29%5E%7Bk%2B1%7D%5Ccdot+%5Cfrac%7B%5Cpi%7D%7B12%7D%2B%5Cfrac%7B%5Cpi+k%7D%7B2%7D%5C%3B+%2Ck%5Cin+Z)
Если из уравнения
![cos^2(\frac{5\pi}{4}-x)=\frac{1}{4}](https://tex.z-dn.net/?f=cos%5E2%28%5Cfrac%7B5%5Cpi%7D%7B4%7D-x%29%3D%5Cfrac%7B1%7D%7B4%7D)
получить при извлечении квадратного корня два случая:
![cos(\frac{5\pi }{4}-x)=\pm \frac{1}{2}](https://tex.z-dn.net/?f=cos%28%5Cfrac%7B5%5Cpi+%7D%7B4%7D-x%29%3D%5Cpm+%5Cfrac%7B1%7D%7B2%7D)
, то надо потом посмотреть, какие решения накладываются друг на друга , и отбросить ненужное. Поэтому всегда лучше пользоваться формулой понижения степени, которой мы уже воспользовались, заменив квадрат косинуса на дробь.
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