Đáp án:
\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{3\pi }}{{14}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k\pi \\
x = k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\sin 2x + \sin x = 0\\
\Leftrightarrow 2\sin x.\cos x + \sin x = 0\\
\Leftrightarrow \sin x.\left( {2\cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
2\cos x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = - \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = \cos \dfrac{{2\pi }}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sin x + \cos 2x - 1 = 0\\
\Leftrightarrow \sin x + \left( {1 - 2{{\sin }^2}x} \right) - 1 = 0\\
\Leftrightarrow \sin x + 1 - 2{\sin ^2}x - 1 = 0\\
\Leftrightarrow \sin x - 2{\sin ^2}x = 0\\
\Leftrightarrow \sin x.\left( {1 - 2\sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
1 - 2\sin x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \sin \dfrac{\pi }{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \left( {\pi - \dfrac{\pi }{6}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\cos x + \cos 2x + 1 = 0\\
\Leftrightarrow \cos x + \left( {2{{\cos }^2}x - 1} \right) + 1 = 0\\
\Leftrightarrow \cos x + 2{\cos ^2}x - 1 + 1 = 0\\
\Leftrightarrow \cos x + 2{\cos ^2}x = 0\\
\Leftrightarrow \cos x.\left( {1 + 2\cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
1 + 2\cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = - \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \cos \dfrac{{2\pi }}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
{\sin ^2}3x + {\sin ^2}4x = 1\\
\Leftrightarrow {\sin ^2}3x + \left( {1 - {{\cos }^2}4x} \right) = 1\\
\Leftrightarrow {\sin ^2}3x + 1 - {\cos ^2}4x - 1 = 0\\
\Leftrightarrow {\sin ^2}3x - {\cos ^2}4x = 0\\
\Leftrightarrow \left( {\sin 3x - \cos x4x} \right).\left( {\sin 3x + \cos 4x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 3x - \cos 4x = 0\\
\sin 3x + \cos 4x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 3x = \cos 4x\\
\sin 3x = - \cos 4x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 3x = \sin \left( {\dfrac{\pi }{2} - 4x} \right)\\
\sin 3x = - \sin \left( {\dfrac{\pi }{2} - 4x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 3x = \sin \left( {\dfrac{\pi }{2} - 4x} \right)\\
\sin 3x = \sin \left( {4x - \dfrac{\pi }{2}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} - 4x + k2\pi \\
3x = \pi - \left( {\dfrac{\pi }{2} - 4x} \right) + k2\pi \\
3x = 4x - \dfrac{\pi }{2} + k2\pi \\
3x = \pi - \left( {4x - \dfrac{\pi }{2}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} - 4x + k2\pi \\
3x = \dfrac{\pi }{2} + 4x + k2\pi \\
3x = 4x - \dfrac{\pi }{2} + k2\pi \\
3x = \dfrac{{3\pi }}{2} - 4x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
7x = \dfrac{\pi }{2} + k2\pi \\
- x = \dfrac{\pi }{2} + k2\pi \\
- x = - \dfrac{\pi }{2} + k2\pi \\
7x = \dfrac{{3\pi }}{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
x = - \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{{3\pi }}{{14}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{3\pi }}{{14}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\sin \left( {\dfrac{\pi }{6} + x} \right) + \cos \left( {\dfrac{\pi }{3} + 2x} \right) = 1\\
\Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) + \cos \left[ {2.\left( {\dfrac{\pi }{6} + x} \right)} \right] = 1\\
\Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) + \left[ {1 - 2{{\sin }^2}\left( {\dfrac{\pi }{6} + x} \right)} \right] = 1\\
\Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) + 1 - 2{\sin ^2}\left( {\dfrac{\pi }{6} + x} \right) = 1\\
\Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) - 2{\sin ^2}\left( {\dfrac{\pi }{6} + x} \right) = 0\\
\Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right)\left( {1 - 2\sin \left( {\dfrac{\pi }{6} + x} \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {\dfrac{\pi }{6} + x} \right) = 0\\
1 - 2\sin \left( {\dfrac{\pi }{6} + x} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {\dfrac{\pi }{6} + x} \right) = 0\\
\sin \left( {\dfrac{\pi }{6} + x} \right) = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {\dfrac{\pi }{6} + x} \right) = 0\\
\sin \left( {\dfrac{\pi }{6} + x} \right) = \sin \dfrac{\pi }{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{6} + x = k\pi \\
\dfrac{\pi }{6} + x = \dfrac{\pi }{6} + k2\pi \\
\dfrac{\pi }{6} + x = \left( {\pi - \dfrac{\pi }{6}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k\pi \\
x = k2\pi \\
x = \pi - \dfrac{\pi }{6} - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k\pi \\
x = k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
