cos(2x+ $\frac{1}{3}$ $\pi$) + sin(2x+$\frac{1}{3}$ $\pi$) = $\frac{1}{2}$ $\sqrt{6}$

Đáp án:

\(\left[ \begin{array}{l}
x =  – \dfrac{\pi }{8} + k\pi \\
x = \dfrac{\pi }{{24}} + k\pi 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
\cos \left( {2x + \dfrac{\pi }{3}} \right) + \sin \left( {2x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 6 }}{2}\\
 \to \dfrac{1}{{\sqrt 2 }}.\cos \left( {2x + \dfrac{\pi }{3}} \right) + \dfrac{1}{{\sqrt 2 }}.\sin \left( {2x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \to \sin \dfrac{\pi }{4}.\cos \left( {2x + \dfrac{\pi }{3}} \right) + \cos \dfrac{\pi }{4}.\sin \left( {2x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \to \sin \left( {\dfrac{\pi }{4} + 2x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \to \left[ \begin{array}{l}
2x + \dfrac{{7\pi }}{{12}} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{{7\pi }}{{12}} = \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x =  – \dfrac{\pi }{4} + k2\pi \\
2x = \dfrac{\pi }{{12}} + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  – \dfrac{\pi }{8} + k\pi \\
x = \dfrac{\pi }{{24}} + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)