Đáp án:2)\[\alpha = \pi - \arcsin (\frac{{ - 9}}{{11}})\]
4)\[\alpha {\rm{ = }}{\mathop{\rm arccot}\nolimits} ( - 3) + k2\pi \]
Giải thích các bước giải:
\[\begin{array}{l}
2)\sin \alpha = \frac{{ - 9}}{{11}}(\pi < \alpha < \frac{{3\pi }}{2})\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\alpha = \arcsin (\frac{{ - 9}}{{11}}) + k2\pi }\\
{\alpha = \pi - \arcsin (\frac{{ - 9}}{{11}}) + k2\pi }
\end{array}} \right.\\
\Leftrightarrow \alpha = \pi - \arcsin (\frac{{ - 9}}{{11}})\\
4)\cot \alpha = - 3(\frac{{3\pi }}{2} < \alpha < 2\pi )\\
\Leftrightarrow \alpha = {\rm{arccot( - 3) + k}}\pi \\
\Leftrightarrow \alpha {\rm{ = }}{\mathop{\rm arccot}\nolimits} ( - 3) + k2\pi
\end{array}\]
