Question

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

Answer 1

1) Умножим обе части уравнения на 1/2:

$\frac{1}{2}sinx+\frac{\sqrt3}{2}cosx=sin2x \\\ cos\frac{\pi}{3}sinx+sin\frac{\pi}{3}cosx=sin2x \\\ sin(x+\frac{\pi}{3})-sin2x=0 \\\ 2sin\frac{x+\frac{\pi}{3}-2x}{2}cos\frac{x+\frac{\pi}{3}+2x}{2}=0 \\\ -sin(\frac{x}{2}-\frac{\pi}{6})cos(\frac{3x}{2}+\frac{\pi}{6})=0$

$sin(\frac{x}{2}-\frac{\pi}{6}) = 0$ или $cos(\frac{3x}{2}+\frac{\pi}{6})=0$

$\frac{x}{2}-\frac{\pi}{6} = \pi k$ или $\frac{3x}{2}+\frac{\pi}{6}=\frac{\pi}{2}+\pi n$

$\frac{x}{2}=\frac{\pi}{6} + \pi k \\\ x=\frac{\pi}{3} + 2\pi k$ или $\frac{3x}{2}+\frac{\pi}{6}=\frac{\pi}{2}+\pi n \\\ \frac{3x}{2}=\frac{\pi}{2}-\frac{\pi}{6}+\pi n \\\ x=\frac{2\pi}{9}+\frac{2\pi n}{3}$

2) Умножим обе части уравнения на 1/2:

$\frac{1}{2}cos3x+\frac{\sqrt3}{2}sin3x=-\frac{1}{2} \\\ sin(3x+\frac{\pi}{6})=-\frac{1}{2} \\\ 3x+\frac{\pi}{6}=(-1)^{k+1}\frac{\pi}{6}+\pi k \\\ 3x=-\frac{\pi}{6}+(-1)^{k+1}\frac{\pi}{6}+\pi k \\\ x=-\frac{\pi}{18}+(-1)^{k+1}\frac{\pi}{18}+\frac{\pi k}{3}$