Đáp án:
$x = \frac{\pi }{3} + k2\pi $
$x = \pi + k2\pi = k\pi $
Giải thích các bước giải:
\[\begin{array}{l}
a.\sqrt {1 + 3} .\sin \left( {x - \frac{\pi }{6}} \right) = 1\\
\Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \frac{\pi }{6} = \frac{\pi }{6} + k2\pi \\
x - \frac{\pi }{6} = \frac{{5\pi }}{6} + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{3} + k2\pi \\
x = \pi + k2\pi = k\pi
\end{array} \right.\\
b.\sqrt {3 + 1} .\sin \left( {2x - \frac{\pi }{3}} \right) = \sqrt 2 \\
\Leftrightarrow \sin \left( {2x - \frac{\pi }{3}} \right) = \frac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\
2x - \frac{\pi }{3} = \frac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{7\pi }}{{24}} + k\pi \\
x = \frac{{13}}{{24}} + k\pi
\end{array} \right.
\end{array}\]
