1) sinx+√3cosx=2sin2x       2)cos3x+√3sin3x=-1 

1) Умножим обе части уравнения на 1/2: [latex]\frac{1}{2}sinx+\frac{\sqrt3}{2}cosx=sin2x \\\ cos\frac{\pi}{3}sinx+sin\frac{\pi}{3}cosx=sin2x \\\ sin(x+\frac{\pi}{3})-sin2x=0 \\\ 2sin\frac{x+\frac{\pi}{3}-2x}{2}cos\frac{x+\frac{\pi}{3}+2x}{2}=0 \\\ -sin(\frac{x}{2}-\frac{\pi}{6})cos(\frac{3x}{2}+\frac{\pi}{6})=0[/latex] [latex]sin(\frac{x}{2}-\frac{\pi}{6}) = 0[/latex] или [latex]cos(\frac{3x}{2}+\frac{\pi}{6})=0[/latex] [latex]\frac{x}{2}-\frac{\pi}{6} = \pi k[/latex] или [latex]\frac{3x}{2}+\frac{\pi}{6}=\frac{\pi}{2}+\pi n[/latex] [latex]\frac{x}{2}=\frac{\pi}{6} + \pi k \\\ x=\frac{\pi}{3} + 2\pi k [/latex] или [latex]\frac{3x}{2}+\frac{\pi}{6}=\frac{\pi}{2}+\pi n \\\ \frac{3x}{2}=\frac{\pi}{2}-\frac{\pi}{6}+\pi n \\\ x=\frac{2\pi}{9}+\frac{2\pi n}{3}[/latex]   2) Умножим обе части уравнения на 1/2: [latex]\frac{1}{2}cos3x+\frac{\sqrt3}{2}sin3x=-\frac{1}{2} \\\ sin(3x+\frac{\pi}{6})=-\frac{1}{2} \\\ 3x+\frac{\pi}{6}=(-1)^{k+1}\frac{\pi}{6}+\pi k \\\ 3x=-\frac{\pi}{6}+(-1)^{k+1}\frac{\pi}{6}+\pi k \\\ x=-\frac{\pi}{18}+(-1)^{k+1}\frac{\pi}{18}+\frac{\pi k}{3} [/latex]