Đáp án:
$\begin{array}{l}
1)a)\sin 3x = - \frac{{\sqrt 3 }}{2}\\
\Rightarrow \left[ \begin{array}{l}
3x = - \frac{\pi }{3} + k2\pi \\
3x = \pi + \frac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{ - \pi }}{9} + \frac{{k2\pi }}{3}\\
x = \frac{{4\pi }}{9} + \frac{{k2\pi }}{9}
\end{array} \right.\left( {k \in Z} \right)\\
b)\sin 6x + \sqrt 3 \cos 6x = - \sqrt 3 \\
\Rightarrow \frac{1}{2}.\sin 6x + \frac{{\sqrt 3 }}{2}\cos 6x = - \frac{{\sqrt 3 }}{2}\\
\Rightarrow \sin \frac{\pi }{6}.\sin 6x + \cos \frac{\pi }{6}.\sin 6x = - \frac{{\sqrt 3 }}{2}\\
\Rightarrow \cos \left( {6x - \frac{\pi }{6}} \right) = \cos \frac{{5\pi }}{6}\\
\Rightarrow \left[ \begin{array}{l}
6x - \frac{\pi }{6} = \frac{{5\pi }}{6} + k2\pi \\
6x - \frac{\pi }{6} = \frac{{ - 5\pi }}{6} + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + \frac{{k\pi }}{3}\\
x = - \frac{\pi }{9} + \frac{{k\pi }}{3}
\end{array} \right.\\
5)\\
a)\,{u_1} = - 3;{u_2} = - \frac{5}{4};{u_3} = - 1;{u_4} = - \frac{9}{{10}};{u_5} = \frac{{ - 11}}{{13}}
\end{array}$
b) Dãy số tăng
