Sin2x + 1 = Cosx + 2Sinx
Sin2x - 2Sinx + 1 - Cosx = 0
(2SinxCosx - 2Sinx) + (1 - Cosx) = 0
2Sinx(Cosx - 1) - (Cosx - 1) = 0
(Cosx - 1)(2Sinx - 1) = 0
![\left[\begin{array}{ccc}Cosx-1=0\\2Sinx-1=0\end{array}\right\\\\\\\left[\begin{array}{ccc}Cosx=1\\Sinx=\frac{1}{2} \end{array}\right\\\\\\\left[\begin{array}{ccc}x=2\pi n,n\in Z \\x=(-1)^{n}arcSin\frac{1}{2}+\pi n,n\in Z \end{array}\right\\\\\\\left[\begin{array}{ccc}x=2\pi n,n\jn Z \\x=(-1)^{n}\frac{\pi }{6}+\pi n,n\in Z \end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DCosx-1%3D0%5C%5C2Sinx-1%3D0%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DCosx%3D1%5C%5CSinx%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%3D2%5Cpi%20n%2Cn%5Cin%20Z%20%5C%5Cx%3D%28-1%29%5E%7Bn%7DarcSin%5Cfrac%7B1%7D%7B2%7D%2B%5Cpi%20n%2Cn%5Cin%20Z%20%20%20%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%3D2%5Cpi%20n%2Cn%5Cjn%20Z%20%5C%5Cx%3D%28-1%29%5E%7Bn%7D%5Cfrac%7B%5Cpi%20%7D%7B6%7D%2B%5Cpi%20n%2Cn%5Cin%20Z%20%20%20%5Cend%7Barray%7D%5Cright)
1)x^2 +2x+x+2 = 2x^2 -4x-x+2
-x^2 + 8x = 0
x^2 - 8x=0
2) 4x^2 - 6x^2 -2x = 5
2x^2 +2x +5 =0
3) 3x^2 - 2x +9x - 6= 8x^2 - 12 + 10x - 15
5x^2 +3x - 9= 0
4) 9x^2 + 4 +12x = x^2 - 3x +2x-6
8x^2 +13x - 2= 0
5) x^2 + 1 - 3x -x + 3x^2 = x
4x^2 - 5x + 1=0
(5a+6a+3a)-10a=14a-10a=4a
Х² - 6х - 7 = 0
х² + х - 7х - 7 = 0
х(х+1) - 7(х+1) = 0
(х+1)(х-7) = 0
х+1 = 0 или х-7 = 0
х = -1 х = 7
Ответ: х=-1; х=7