Ответ :

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

[tex]\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[/tex]

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