Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sin x = \dfrac{1}{2}\\
\Leftrightarrow \sin x = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \pi - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
b,\\
\cos x = 0\\
\Leftrightarrow \cos x = \cos \dfrac{\pi }{2}\\
\Leftrightarrow x = \pm \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
c,\\
\sin 2x = - 1\\
\Leftrightarrow \sin 2x = \sin \left( { - \dfrac{\pi }{2}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{ - \pi }}{2} + k2\pi \\
2x = \pi - \dfrac{{ - \pi }}{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow 2x = \dfrac{{ - \pi }}{2} + k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{4} + k\pi \\
d,\\
2\tan 2x = 1\\
\Leftrightarrow \tan 2x = \dfrac{1}{2}\\
\Leftrightarrow 2x = \arctan \dfrac{1}{2} + k\pi \\
\Leftrightarrow x = \dfrac{{\arctan \dfrac{1}{2}}}{2} + \dfrac{{k\pi }}{2}
\end{array}\)
