Đáp án:
\(\begin{array}{l}
c,\\
\left[ \begin{array}{l}
x = \dfrac{1}{2} + k\pi \\
x = \dfrac{{\pi + 1}}{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
x = \dfrac{{5\pi }}{9} + \dfrac{{k4\pi }}{3}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
{\cos ^2}\left( {2x - 1} \right) = 1\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \left( {2x - 1} \right) = 1\\
\cos \left( {2x - 1} \right) = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = k2\pi \\
2x - 1 = \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 1 + k2\pi \\
2x = \pi + 1 + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2} + k\pi \\
x = \dfrac{{\pi + 1}}{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
\sin \left( {\dfrac{{3x}}{2} - \dfrac{\pi }{3}} \right) = 1\\
\Leftrightarrow \dfrac{{3x}}{2} - \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow \dfrac{{3x}}{2} = \dfrac{\pi }{2} + \dfrac{\pi }{3} + k2\pi \\
\Leftrightarrow \dfrac{{3x}}{2} = \dfrac{{5\pi }}{6} + k2\pi \\
\Leftrightarrow 3x = \dfrac{{5\pi }}{3} + k4\pi \\
\Leftrightarrow x = \dfrac{{5\pi }}{9} + \dfrac{{k4\pi }}{3}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
