sin(2x+π/3)-cos(x-π/4)=0

Đáp án: $\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{ - \pi }}{{12}} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)$

Giải thích các bước giải:

$\begin{array}{l}
\sin \left( {2x + \dfrac{\pi }{3}} \right) - \cos \left( {x - \dfrac{\pi }{4}} \right) = 0\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \cos \left( {x - \dfrac{\pi }{4}} \right)\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{2} - x + \dfrac{\pi }{4}} \right)\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{{3\pi }}{4} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} - x + k2\pi \\
2x + \dfrac{\pi }{3} = \pi  - \dfrac{{3\pi }}{4} + x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{{5\pi }}{{12}} + k2\pi \\
x = \dfrac{{ - \pi }}{{12}} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{ - \pi }}{{12}} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
Vay\,\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{ - \pi }}{{12}} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$