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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)
2 log7 27 - log7 81 -2log7 21 = 2 log7 3^3 - log7 3^4 - 2log7 7 - 2 log7 3= 6 log7 3 - 4 log7 3 -2 log7 7 - 2 log7 3= -2
Первое ну я так думаю потому что 2 не может быть 6
Сторона квадрата - это диаметр= 32
S = 32² = 1024