Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a) x=\ \frac{\prod }{18} +\frac{k2\prod }{3}\\ hoặc\ x=\ \frac{5\prod }{18} \ +\frac{k2\prod }{3}\\ b) \ x=\frac{\prod }{3} +k\prod \\ hoặc\ x=\ \frac{\prod }{6} \ +\ k\prod \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a) \ sin3x=sin\frac{\prod }{6}\\ \Leftrightarrow \ 3x=\frac{\prod }{6} \ +k2\prod \\ hoặc\ 3x=\ \frac{5\prod }{6} \ +k2\prod \\ \Leftrightarrow x=\ \frac{\prod }{18} +\frac{k2\prod }{3}\\ hoặc\ x=\ \frac{5\prod }{18} \ +\frac{k2\prod }{3}\\ \\ \\ b) cos\left( 2x-\frac{\prod }{2}\right) =\frac{\sqrt{3}}{2}\\ \Leftrightarrow cos\left( 2x-\frac{\prod }{2}\right) =\ cos\frac{\prod }{6}\\ \Leftrightarrow 2x-\frac{\prod }{2} =\frac{\prod }{6} \ +\ k2\prod \\ hoặc\ 2x-\frac{\prod }{2} =-\frac{\prod }{6} \ +\ k2\prod \\ \Leftrightarrow x=\frac{\prod }{3} +k\prod \\ hoặc\ x=\ \frac{\prod }{6} \ +\ k\prod \end{array}$
