Giải các phương trình sau:a) \(\sin 3x + \cos 3x = 1\)b) \(2\sin x + \cos x - \sin 2x - 1 = 0\)

* Hướng dẫn giải

a) \(\sin 3x + \cos 3x = \sqrt 2 \sin x \Leftrightarrow \sqrt 2 \sin \left( {3x + \frac{\pi }{4}} \right) = \sqrt 2 \sin x\)

\(\begin{array}{l}
 \Leftrightarrow \left[ \begin{array}{l}
3x + \frac{\pi }{4} = x + k2\pi \\
3x + \frac{\pi }{4} = \pi  - x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{\pi }{8} + k\pi \\
x = \frac{{3\pi }}{{16}} + k\frac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

b) \(2\sin x + \cos x - \sin 2x - 1 = 0 \Leftrightarrow \left( {2\sin x - 1} \right)\left( {1 - \cos x} \right) = 0\)

\(\begin{array}{l}
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \frac{1}{2}\\
\cos x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi \\
x = k2\pi 
\end{array} \right.,\left( {k \in Z} \right)
\end{array}\)