Resposta:

[tex]S=\left\{\dfrac{\pi}{18},\dfrac{5\pi}{18},\dfrac{13\pi}{18},\dfrac{17\pi}{18}\right\}[/tex]

Explicação passo a passo:

[tex]sen(3x)=\dfrac{1}{2}\\ arcsen(sen(3x))=arcsen\left(\dfrac{1}{2}\right)\\\\3x=2\pi k+\dfrac{\pi}{6}\qquad\text{ou} \qquad 3x=2\pi k+\dfrac{5\pi}{6} \qquad |\ k\in\mathbb{Z}[/tex]

[tex]x=\dfrac{2\pi k}{3} +\dfrac{\pi}{18}\qquad\text{ou} \qquad x=\dfrac{2\pi k}{3}+\dfrac{5\pi}{18} \qquad |\ k\in\mathbb{Z}[/tex]

[tex]x=\dfrac{12\pi k}{18} +\dfrac{\pi}{18}\qquad\text{ou} \qquad x=\dfrac{12\pi k}{18}+\dfrac{5\pi}{18} \qquad |\ k\in\mathbb{Z}[/tex]

Para [tex]k = 0[/tex] :

[tex]x = \dfrac{\pi}{18}\qquad \text{ou} \qquad x=\dfrac{5\pi}{18}[/tex]

Para [tex]k = 1[/tex] :

[tex]x = \dfrac{13\pi}{18}\qquad \text{ou} \qquad x=\dfrac{17\pi}{18}[/tex]

Para [tex]k = 2[/tex] :

[tex]x = \dfrac{25\pi}{18}\qquad \text{ou} \qquad x=\dfrac{29\pi}{18}[/tex]

Percebe-se que para [tex]k\geq2[/tex], [tex]x > \pi[/tex]. Portanto:

[tex]S=\left\{\dfrac{\pi}{18},\dfrac{5\pi}{18},\dfrac{13\pi}{18},\dfrac{17\pi}{18}\right\}[/tex]