Đáp án:
2) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1){\sin ^2}x = \dfrac{1}{4}\\
\to \left[ \begin{array}{l}
\sin x = \dfrac{1}{2}\\
\sin x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
2)\left[ \begin{array}{l}
\sin x - 1 = 0\\
2\sin x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
