График функции y=5-|x+8| это график функции y=-|x|, сдвинутый на 8 влево и поднятый на 5.
Так как ОЗФ у функции y=-|x|
![E(f)=(- \infty ; 0 ]](https://tex.z-dn.net/?f=E%28f%29%3D%28-+%5Cinfty+%3B+0+%5D)
то ОЗФ у функции y=5-|x+8|
![E(f)=(- \infty; 5]](https://tex.z-dn.net/?f=E%28f%29%3D%28-+%5Cinfty%3B+5%5D)
Ответ:
![E(f)=(- \infty; 5]](https://tex.z-dn.net/?f=E%28f%29%3D%28-+%5Cinfty%3B+5%5D)
Положим
![\arcsin x=t~~~~\Rightarrow~~~~ x=\sin t](https://tex.z-dn.net/?f=%5Carcsin+x%3Dt~~~~%5CRightarrow~~~~+x%3D%5Csin+t)
и при этом
![t\in \bigg[- \dfrac{\pi}{2} ;\dfrac{\pi}{2} \bigg].](https://tex.z-dn.net/?f=t%5Cin+%5Cbigg%5B-+%5Cdfrac%7B%5Cpi%7D%7B2%7D+%3B%5Cdfrac%7B%5Cpi%7D%7B2%7D+%5Cbigg%5D.)
![12t^2+2 \pi t+6 \pi \arccos\bigg(\sin t\bigg)=4 \pi ^2\\ \\ 12t^2+2\pi t+6\pi\arccos\bigg(\cos\bigg( \dfrac{\pi}{2}-t\bigg)\bigg)=4\pi^2\\ \\ 12t^2+2 \pi t+6\pi\bigg( \dfrac{\pi}{2}-t\bigg)=4\pi^2\\ \\ 12t^2+2\pi t+3\pi^2-6\pi t=4\pi^2\\ \\ 12t^2-4\pi t-\pi^2 =0\\ \\ D=16\pi^2+48\pi^2=64\pi^2;~~~~ \sqrt{D} =8\pi\\ \\ t_1= \dfrac{\pi}{2} ;~~~~~~~~ t_2=-\dfrac{\pi}{6}](https://tex.z-dn.net/?f=12t%5E2%2B2+%5Cpi+t%2B6+%5Cpi+%5Carccos%5Cbigg%28%5Csin+t%5Cbigg%29%3D4+%5Cpi+%5E2%5C%5C+%5C%5C+12t%5E2%2B2%5Cpi+t%2B6%5Cpi%5Carccos%5Cbigg%28%5Ccos%5Cbigg%28+%5Cdfrac%7B%5Cpi%7D%7B2%7D-t%5Cbigg%29%5Cbigg%29%3D4%5Cpi%5E2%5C%5C+%5C%5C+12t%5E2%2B2+%5Cpi+t%2B6%5Cpi%5Cbigg%28+%5Cdfrac%7B%5Cpi%7D%7B2%7D-t%5Cbigg%29%3D4%5Cpi%5E2%5C%5C+%5C%5C+12t%5E2%2B2%5Cpi+t%2B3%5Cpi%5E2-6%5Cpi+t%3D4%5Cpi%5E2%5C%5C+%5C%5C+12t%5E2-4%5Cpi+t-%5Cpi%5E2++%3D0%5C%5C+%5C%5C+D%3D16%5Cpi%5E2%2B48%5Cpi%5E2%3D64%5Cpi%5E2%3B~~~~+%5Csqrt%7BD%7D+%3D8%5Cpi%5C%5C+%5C%5C+t_1%3D+%5Cdfrac%7B%5Cpi%7D%7B2%7D+%3B~~~~~~~~+t_2%3D-%5Cdfrac%7B%5Cpi%7D%7B6%7D+)
Осуществив обратную замену, находим
![x_1=\sin \dfrac{\pi}{2} =1;~~~~~ x_2=\sin\bigg(-\dfrac{\pi}{6} \bigg)=- \dfrac{1}{2}](https://tex.z-dn.net/?f=x_1%3D%5Csin+%5Cdfrac%7B%5Cpi%7D%7B2%7D+%3D1%3B~~~~~+x_2%3D%5Csin%5Cbigg%28-%5Cdfrac%7B%5Cpi%7D%7B6%7D+%5Cbigg%29%3D-+%5Cdfrac%7B1%7D%7B2%7D+)
Всё расписал на листе. 20шкпари хшцптм9шцпт0штцам0щт
D=2014^2-4·1·(-2015)=4056196+4·2015=√4064256=2016
x1=(-2014+2016)/2=1
x2=(-2014-2016)/2=-2015
Первые три системы решила методом сложения