Sin2x+2sinx=2-2cosx     решение с объяснением

[latex]\sin 2x + 2\sin x = 2 - 2\cos x\\\\ 2\sin x\cos x + 2\sin x + 2\cos x = 2\\\\ 1 + 2\sin x\cos x + 2\sin x + 2\cos x = 3\\\\ \left[ \ \sin^2 x + \cos^2 x = 1 \ \right]\\\\ \sin^2 x + 2\sin x\cos x + \cos^2 x + 2\sin x + 2\cos x = 3\\\\ (\sin x + \cos x)^2 + 2(\sin x + \cos x) + 1 = 4[/latex] [latex]\left[ \ a^2 + 2ab + b^2 = (a + b)^2 \ \right]\\\\(\sin x + \cos x + 1)^2 - 4 = 0\\\\\left[ \ a^2 - b^2 = (a - b)(a+ b) \ \right]\\\\(\sin x + \cos x + 1 - 2)(\sin x + \cos x + 1 + 2) = 0\\\\ \left[ \ a*b = 0 \Rightarrow a = 0 \vee b = 0 \ \right][/latex] [latex]\sin x + \cos x - 1 = 0 \ \ \ \ \ \ \ \ \sin x + \cos x + 3 = 0\\\\ \sin x + \cos x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \sin x + \cos x = - 3 \ (!)\\\\ \sin x*\frac{\sqrt{2}}2 + \frac{\sqrt{2}}2*\cos x = \frac{\sqrt{2}}2\\\\ \sin x \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4})\cos x = \frac{\sqrt{2}}2\\\\ \left[ \ \sin a \cos b + \cos a \sin b = \sin(a + b) \ \right]\\\\\sin(x + \frac{\pi}{4}) = \frac{2}{\sqrt{2}}\\\\ [/latex] [latex]x + \frac{\pi}{4} = \frac{\pi}{4} + 2\pi n, \ n \in \mathbb{Z} \ \ \ \ \ x + \frac{\pi}{4} = \frac{3\pi}{4} + 2\pi n, \ n \in \mathbb{Z}\\\\ \boxed{x =2\pi n, \ n \in \mathbb{Z} } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{x = \frac{\pi}{2} + 2\pi n, \ \ n \in \mathbb{Z}} [/latex]