Забей в решение будет лучше чем ждать пока тебе ответят)
3)
16x²+25=0
16x²=-25
x²=-25/16 - корней нет, т.к. справа отрицательное число, слева х²
4)
(3х-5)²-16=0
(3х-5-4)(3х-5+4)=0
(3х-9)(3х-1)=0
3х-9=0
3х=9
х=3
или
3х-1=0
3х=1
х=1/3
![16^{\sin x}-6\cdot 4^{\sin x}+8=0](https://tex.z-dn.net/?f=16%5E%7B%5Csin+x%7D-6%5Ccdot+4%5E%7B%5Csin+x%7D%2B8%3D0)
![(4^2)^{\sin x}-6\cdot 4^{\sin x}+8=0](https://tex.z-dn.net/?f=%284%5E2%29%5E%7B%5Csin+x%7D-6%5Ccdot+4%5E%7B%5Csin+x%7D%2B8%3D0)
Используем свойство степени
![(a^n)^m=a^{n\cdot m}](https://tex.z-dn.net/?f=%28a%5En%29%5Em%3Da%5E%7Bn%5Ccdot+m%7D)
, имеем
![4^{2\sin x}-6\cdot4^{\sin x}+8=0](https://tex.z-dn.net/?f=4%5E%7B2%5Csin+x%7D-6%5Ccdot4%5E%7B%5Csin+x%7D%2B8%3D0)
Пусть
![4^{\sin x}=t](https://tex.z-dn.net/?f=4%5E%7B%5Csin+x%7D%3Dt)
и при этом
![t\ \textgreater \ 0](https://tex.z-dn.net/?f=t%5C+%5Ctextgreater+%5C+0)
. Получаем
![t^2-6t+8=0](https://tex.z-dn.net/?f=t%5E2-6t%2B8%3D0+)
Согласно теореме Виета:
![t_1=2;~~~ t_2=4](https://tex.z-dn.net/?f=t_1%3D2%3B~~~+t_2%3D4)
Обратная замена:
![4^{\sin x}=2;~~\Rightarrow~~~ (2^2)^{\sin x}=2\\ \\ 2^{2\sin x}=2\\ \\ 2\sin x=1\\ \\ \sin x=0.5;~~~\Rightarrow~~~~ \boxed{x_1=(-1)^k\cdot \frac{\pi}{6}+ \pi k,k \in \mathbb{Z} }](https://tex.z-dn.net/?f=4%5E%7B%5Csin+x%7D%3D2%3B~~%5CRightarrow~~~+%282%5E2%29%5E%7B%5Csin+x%7D%3D2%5C%5C+%5C%5C+2%5E%7B2%5Csin+x%7D%3D2%5C%5C+%5C%5C+2%5Csin+x%3D1%5C%5C+%5C%5C+%5Csin+x%3D0.5%3B~~~%5CRightarrow~~~~+%5Cboxed%7Bx_1%3D%28-1%29%5Ek%5Ccdot+%5Cfrac%7B%5Cpi%7D%7B6%7D%2B+%5Cpi+k%2Ck+%5Cin+%5Cmathbb%7BZ%7D+%7D)
![4^{\sin x}=4\\ \\ \sin x=1;~~~\Rightarrow~~~ \boxed{x_2= \frac{\pi}{2}+2 \pi k,k \in \mathbb{Z} }](https://tex.z-dn.net/?f=4%5E%7B%5Csin+x%7D%3D4%5C%5C+%5C%5C+%5Csin+x%3D1%3B~~~%5CRightarrow~~~+%5Cboxed%7Bx_2%3D+%5Cfrac%7B%5Cpi%7D%7B2%7D%2B2+%5Cpi+k%2Ck+%5Cin+%5Cmathbb%7BZ%7D+%7D)