Đáp án:
$a){x = {5^0} + \dfrac{1}{2}\arcsin \left( {\dfrac{3}{5}} \right) + k180^0 ;x = {5^0} + \dfrac{180^0 }{2} - \dfrac{1}{2}\arcsin \left( {\dfrac{3}{5}} \right) + k180^0 \left( {k \in Z} \right)}$
$b)x = \dfrac{1}{3}{.2^0} + \dfrac{1}{3}{\mathop{\rm arccot}\nolimits} 10 + k\dfrac{180^0 }{3}\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
a)\sin \left( {2x - {{10}^0}} \right) = \dfrac{3}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - {10^0} = \arcsin \left( {\dfrac{3}{5}} \right) + k2.180^0 \\
2x - {10^0} = 180^0 - \arcsin \left( {\dfrac{3}{5}} \right) + k2.180^0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = {5^0} + \dfrac{1}{2}\arcsin \left( {\dfrac{3}{5}} \right) + k180^0 \\
x = {5^0} + \dfrac{180^0 }{2} - \dfrac{1}{2}\arcsin \left( {\dfrac{3}{5}} \right) + k180^0
\end{array} \right.\left( {k \in Z} \right)\\
b)\cot \left( {3x - {2^0}} \right) = 10\\
\Leftrightarrow 3x - {2^0} = {\mathop{\rm arccot}\nolimits} 10 + k180^0 \\
\Leftrightarrow x = \dfrac{1}{3}{.2^0} + \dfrac{1}{3}{\mathop{\rm arccot}\nolimits} 10 + k\dfrac{180^0 }{3}\left( {k \in Z} \right)
\end{array}$
