Đáp án:
\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = 50^\circ + k.360^\circ \\
x = 170^\circ + k.360^\circ
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = 2 + \arccos \dfrac{3}{5} + k2\pi \\
x = 2 - \arccos \dfrac{3}{5} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
x = 30^\circ + k.90^\circ \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\sin \left( {x - 20^\circ } \right) = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {x - 20^\circ } \right) = \sin 30^\circ \\
\Leftrightarrow \left[ \begin{array}{l}
x - 20^\circ = 30^\circ + k.360^\circ \\
x - 20^\circ = 180^\circ - 30^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 20^\circ = 30^\circ + k.360^\circ \\
x - 20^\circ = 150^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 50^\circ + k.360^\circ \\
x = 170^\circ + k.360^\circ
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
\cos \left( {x - 2} \right) = \dfrac{3}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = \arccos \dfrac{3}{5} + k2\pi \\
x - 2 = - \arccos \dfrac{3}{5} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2 + \arccos \dfrac{3}{5} + k2\pi \\
x = 2 - \arccos \dfrac{3}{5} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
DKXD:\,\,\,\cos \left( {2x - 15^\circ } \right) \ne 0 \Leftrightarrow 2x - 15^\circ \ne 90^\circ + k.180^\circ \\
\Leftrightarrow 2x \ne 105^\circ + k.180^\circ \Leftrightarrow x \ne \dfrac{{105^\circ }}{2} + k.90^\circ \,\,\,\,\,\left( {k \in Z} \right)\\
\tan \left( {2x - 15^\circ } \right) = 1\\
\Leftrightarrow \tan \left( {2x - 15^\circ } \right) = \tan 45^\circ \\
\Leftrightarrow 2x - 15^\circ = 45^\circ + k.180^\circ \\
\Leftrightarrow 2x = 60^\circ + k.180^\circ \\
\Leftrightarrow x = 30^\circ + k.90^\circ \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
