Giải pt lượng giác sau cos (2x+π/3) + sin(2x+π/3) = √6/2

Đáp án:

$\left[\begin{array}{l}x = -\dfrac{\pi}{8} + k\pi\\x = \dfrac{\pi}{24} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\begin{array}{l}\cos\left(2x + \dfrac{\pi}{3}\right)+ \sin\left(2x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt6}{2}\\ \Leftrightarrow \dfrac{1}{\sqrt2}\cos\left(2x + \dfrac{\pi}{3}\right) + \dfrac{1}{\sqrt2}\sin\left(2x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{3} + \dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{3}\\ \Leftrightarrow \sin\left(2x + \dfrac{7\pi}{12}\right)=\sin\dfrac{\pi}{3}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{7\pi}{12} = \dfrac{\pi}{3} + k2\pi\\2x + \dfrac{7\pi}{12} = \dfrac{2\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = -\dfrac{\pi}{4} + k2\pi\\2x = \dfrac{\pi}{12} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{8} + k\pi\\x = \dfrac{\pi}{24} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ \end{array}$