Đáp án:
\(\begin{array}{l}
21,\\
x = \dfrac{{5\pi }}{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
22,\\
x = \dfrac{\pi }{6} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
23,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
24,\\
x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
21,\\
\sin 2x - \sqrt 3 \cos 2x = 2\\
\Leftrightarrow \dfrac{1}{2}\sin 2x - \dfrac{{\sqrt 3 }}{2}\cos 2x = 1\\
\Leftrightarrow \sin 2x.\cos \dfrac{\pi }{3} - \cos 2x.\sin \dfrac{\pi }{3} = 1\\
\Leftrightarrow \sin \left( {2x - \dfrac{\pi }{3}} \right) = 1\\
\Leftrightarrow 2x - \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 2x = \dfrac{{5\pi }}{6} + k2\pi \\
\Leftrightarrow x = \dfrac{{5\pi }}{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
22,\\
\sin x + \sqrt 3 \cos x = 2\\
\Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = 1\\
\Leftrightarrow \sin x.\cos \dfrac{\pi }{3} + \cos x.\sin \dfrac{\pi }{3} = 1\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = 1\\
\Leftrightarrow x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{6} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
23,\\
\sqrt 3 \sin x - \cos x = 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{1}{2}\\
\Leftrightarrow \sin x.\cos \dfrac{\pi }{6} - \cos x.\sin \dfrac{\pi }{6} = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
24,\\
\sin 3x - \sqrt 3 \cos 3x = - 2\\
\Leftrightarrow \dfrac{1}{2}\sin 3x - \dfrac{{\sqrt 3 }}{2}\cos 3x = - 1\\
\Leftrightarrow \sin 3x.\cos \dfrac{\pi }{3} - \cos 3x.\sin \dfrac{\pi }{3} = - 1\\
\Leftrightarrow \sin \left( {3x - \dfrac{\pi }{3}} \right) = - 1\\
\Leftrightarrow 3x - \dfrac{\pi }{3} = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 3x = - \dfrac{\pi }{6} + k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
