\ k=1; 2\\
k=1\ =>\ x=\frac{\pi}{9}+\frac{\pi}{3}=\frac{4\pi}{9}\\" alt="1)\ 3ctg(3x)-\sqrt3=0; \ x\ [\frac{\pi}{6};\pi]\\
3ctg(3x)-\sqrt3=0\\
3ctg(3x)=\sqrt3\\
ctg(3x)=\frac{\sqrt3}{3}\\
3x=\frac{\pi}{3}+\pi k\\
x=\frac{\pi}{9}+\frac{\pi k}{3};\ x\ [\frac{\pi}{6};\pi]\\
\frac{\pi}{6} \leq \frac{\pi}{9}+\frac{\pi k}{3} \leq \pi\\
\frac{1}{6} \leq \frac{1}{9}+\frac{k}{3} \leq 1\\
\frac{1}{18} \leq \frac{k}{3} \leq \frac{8}{9}\\
\frac{1}{6} \leq k \leq 2 \frac{2}{3}\ =>\ k=1; 2\\
k=1\ =>\ x=\frac{\pi}{9}+\frac{\pi}{3}=\frac{4\pi}{9}\\" align="absmiddle" class="latex-formula">
\ x=\frac{\pi}{9}+\frac{2\pi}{3}=\frac{7\pi}{9}\\
2)\ sin(x-\frac{\pi}{6})=-\frac{\sqrt3}{2};\ x>0 \\
|x-\frac{\pi}{6}=-\frac{\pi}{6}+2\pi k\\
|x-\frac{\pi}{6}=-\frac{5\pi}{6}+2\pi k\\
\\
|x=2\pi k\\
|x=-\frac{2\pi}{3}+2\pi k\\
x>0\\
\\
|2\pi k>0 \ =>\ x=2\pi>0(k=0 \ =>\ x=0)\\
|-\frac{2\pi}{3}+2\pi k >0\ =>k=0; x=-\frac{2\pi}{3}<0\ =>\ k=1;\ x=-\frac{2\pi}{3}+\\
+2\pi=\frac{4\pi}{3}>0\\
\frac{4\pi}{3}<2\pi\\" alt="k=2\ =>\ x=\frac{\pi}{9}+\frac{2\pi}{3}=\frac{7\pi}{9}\\
2)\ sin(x-\frac{\pi}{6})=-\frac{\sqrt3}{2};\ x>0 \\
|x-\frac{\pi}{6}=-\frac{\pi}{6}+2\pi k\\
|x-\frac{\pi}{6}=-\frac{5\pi}{6}+2\pi k\\
\\
|x=2\pi k\\
|x=-\frac{2\pi}{3}+2\pi k\\
x>0\\
\\
|2\pi k>0 \ =>\ x=2\pi>0(k=0 \ =>\ x=0)\\
|-\frac{2\pi}{3}+2\pi k >0\ =>k=0; x=-\frac{2\pi}{3}<0\ =>\ k=1;\ x=-\frac{2\pi}{3}+\\
+2\pi=\frac{4\pi}{3}>0\\
\frac{4\pi}{3}<2\pi\\" align="absmiddle" class="latex-formula">
Ответ:
