Решите уравнение sin2x+1=sinx+cosx.

[latex]Sin(2x)+1=Sin(x)+Cos(x) \\ 2Sin(x)Cos(x) + Sin^2(x) + Cos^2(x) - Sin(x) - Cos(x) = 0 \\ (Sin^2(x) + 2Sin(x)Cos(x) + Cos^2(x)) - (Sin(x)+Cos(x)) = 0 \\ (Sin(x) + Cos(x))^2 - (Sin(x)+Cos(x)) = 0 \\ (Sin(x) + Cos(x))(Sin(x) + Cos(x) - 1) = 0 \\ Sin(x) + Cos(x) = 0 \ \ \ \ \ \ \ \ Sin(x) + Cos(x) -1 = 0\\ \sqrt{2}Sin(x+\frac{\pi}{4}) = 0 \ \ \ \ \ \ \ \ \ \ \ \sqrt{2}Sin(x+\frac{\pi}{4}) = 1 \\ Sin(x+\frac{\pi}{4}) = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Sin(x+\frac{\pi}{4}) = \frac{\sqrt{2}}{2}[/latex] [latex]x + \frac{\pi}{4} = \pi n \ \ \ \ x + \frac{\pi}{4} = \frac{\pi}{4}+2\pi m \ \ \ \ x + \frac{\pi}{4} =\frac{3\pi}{4} + 2\pi k \\ x = - \frac{\pi}{4} + \pi n \ \ \ \ \ \ x = 2\pi m \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{2} + 2\pi k[/latex]