Rozwiązanie:
1)
[tex]sin^{2}x-2sinx\geq 0\\t=sinx \wedge t \in <-1,1>\\t^{2}-2t\geq 0\\t(t-2)\geq 0\\t \in <-1,0> \iff sinx \in <-1,0>\\sinx \geq -1 \wedge sinx\leq 0\\x \in <-\pi +2k\pi , 2k\pi >\\k \in \mathbb{Z}[/tex]
2)
[tex]tg3x=\frac{\sqrt{3} }{3} \\3x=\frac{\pi }{6} +k\pi \\x=\frac{\pi}{18} +\frac{k\pi}{3}\\k \in \mathbb{Z}[/tex]
