Odpowiedź:
[tex]x\in\{-\frac{\pi}{2},-\frac{5\pi}{18},-\frac{\pi}{6},\frac{\pi}{18},\frac{\pi}{6},\frac{7\pi}{18},\frac{\pi}{2},\frac{13\pi}{18},\frac{5\pi}{6}\}[/tex]
Szczegółowe wyjaśnienie:
[tex]\left|\text{tg}\left(3x-\frac{\pi}{3}\right)\right|=\frac{\sqrt3}{3}\qquad x\in\left < -\frac{\pi}{2},\pi\right >[/tex]
Najpierw znajdźmy dziedzinę równania.
[tex]3x-\frac{\pi}{3}\neq \frac{\pi}{2}+k\pi,\ k\in\mathbb{Z}\\3x\neq \frac{\pi}{2}+\frac{\pi}{3}+k\pi,\ k\in\mathbb{Z}\\3x\neq \frac{3\pi}{6}+\frac{2\pi}{6}+k\pi,\ k\in\mathbb{Z}\\3x\neq \frac{5\pi}{6}+k\pi\ |:3,\ k\in\mathbb{Z}\\x\neq \frac{5\pi}{18}+\frac{k\pi}{3},\ k\in\mathbb{Z}\\D=\mathbb{R}-\{\frac{5\pi}{18}+\frac{k\pi}{3},\ k\in\mathbb{Z}\}[/tex]
Przejdźmy do rozwiązania równania.
[tex]\left|\text{tg}\left(3x-\frac{\pi}{3}\right)\right|=\frac{\sqrt3}{3}\\\text{tg}\left(3x-\frac{\pi}{3}\right)=\frac{\sqrt3}{3}\vee\text{tg}\left(3x-\frac{\pi}{3}\right)=-\frac{\sqrt3}{3}\\3x-\frac{\pi}{3}=\frac{\pi}{6}+k\pi\vee3x-\frac{\pi}{3}=-\frac{\pi}{6}+k\pi,k\in\mathbb{Z}\\3x=\frac{\pi}{6}+\frac{\pi}{3}+k\pi\vee3x=-\frac{\pi}{6}+\frac{\pi}{3}+k\pi,k\in\mathbb{Z}\\3x=\frac{\pi}{6}+\frac{2\pi}{6}+k\pi\vee3x=-\frac{\pi}{6}+\frac{2\pi}{6}+k\pi,k\in\mathbb{Z}[/tex]
[tex]3x=\frac{3\pi}{6}+k\pi\vee3x=\frac{\pi}{6}+k\pi,k\in\mathbb{Z}\\3x=\frac{\pi}{2}+k\pi\ |:3\vee3x=\frac{\pi}{6}+k\pi\ |:3,k\in\mathbb{Z}\\x=\frac{\pi}{6}+\frac{k\pi}{3}\vee x=\frac{\pi}{18}+\frac{k\pi}{3},k\in\mathbb{Z}[/tex]
Teraz należy podstawiać poszczególne całkowite wartości k i sprawdzać, czy otrzymane liczby należą do przedziału [tex]\left < -\frac{\pi}{2},\pi\right >[/tex].
[tex]k=-2\\x=\frac{\pi}{6}-\frac{2\pi}{3}=\frac{\pi}{6}-\frac{4\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}\ \text{OK}\\x=\frac{\pi}{18}-\frac{2\pi}{3}=\frac{\pi}{18}-\frac{12\pi}{18}=-\frac{11\pi}{18}\ \text{odrzucamy}\\k=-1\\x=\frac{\pi}{6}-\frac{\pi}{3}=\frac{\pi}{6}-\frac{2\pi}{6}=-\frac{\pi}{6}\ \text{OK}\\x=\frac{\pi}{18}-\frac{\pi}{3}=\frac{\pi}{18}-\frac{6\pi}{18}=-\frac{5\pi}{18}\ \text{OK}\\k=0\\x=\frac{\pi}{6}\ \text{OK}\\x=\frac{\pi}{18}\ \text{OK}[/tex]
[tex]k=1\\x=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6}+\frac{2\pi}{6}=\frac{3\pi}{6}=\frac{\pi}{2}\ \text{OK}\\x=\frac{\pi}{18}+\frac{\pi}{3}=\frac{\pi}{18}+\frac{6\pi}{18}=\frac{7\pi}{18}\ \text{OK}\\k=2\\x=\frac{\pi}{6}+\frac{2\pi}{3}=\frac{\pi}{6}+\frac{4\pi}{6}=\frac{5\pi}{6}\ \text{OK}\\x=\frac{\pi}{18}+\frac{2\pi}{3}=\frac{\pi}{18}+\frac{12\pi}{18}=\frac{13\pi}{18}\ \text{OK}[/tex]
Ostatecznie
[tex]x\in\{-\frac{\pi}{2},-\frac{5\pi}{18},-\frac{\pi}{6},\frac{\pi}{18},\frac{\pi}{6},\frac{7\pi}{18},\frac{\pi}{2},\frac{13\pi}{18},\frac{5\pi}{6}\}[/tex]