Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\sin 5x + \sin 9x + 2{\sin ^2}x - 1 = 0\\
\Leftrightarrow 2.\sin \dfrac{{5x + 9x}}{2}.\cos \dfrac{{5x - 9x}}{2} - \left( {1 - 2{{\sin }^2}x} \right) = 0\\
\Leftrightarrow 2\sin 7x.\cos \left( { - 2x} \right) - \cos 2x = 0\\
\Leftrightarrow 2\sin 7x.\cos 2x - \cos 2x = 0\\
\Leftrightarrow \cos 2x.\left( {2\sin 7x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sin 7x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
7x = \dfrac{\pi }{6} + k2\pi \\
7x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{{42}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{5\pi }}{{42}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\\
*)\\
\sin \left( {2x + \dfrac{\pi }{6}} \right) - \sin \left( {2x - \dfrac{\pi }{6}} \right) + 1 = \sqrt 3 \sin 2x\\
\Leftrightarrow \left( {\sin 2x.\cos \dfrac{\pi }{6} + \cos 2x.\sin \dfrac{\pi }{6}} \right) - \left( {\sin 2x.\cos \dfrac{\pi }{6} - \cos 2x.sin\dfrac{\pi }{6}} \right) + 1 = \sqrt 3 \sin 2x\\
\Leftrightarrow 2\cos 2x.\sin \dfrac{\pi }{6} + 1 = \sqrt 3 \sin 2x\\
\Leftrightarrow \cos 2x + 1 = \sqrt 3 \sin 2x\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{1}{2}\cos 2x = \dfrac{1}{2}\\
\Leftrightarrow \sin 2x.\cos \dfrac{\pi }{6} - \cos 2x.sin\dfrac{\pi }{6} = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \sin \left( {2x - \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.
\end{array}\)
