Đáp án: $C$
$2sin2x-2cos2x=√2$
Chia cả hai vế cho $\sqrt[]{a^2+b^2}=2√2$
$⇒$ $\dfrac{√2}{2}sin2x-$ $\dfrac{√2}{2}cos2x=0,5$
$⇔$ $sin2x.cos\dfrac{\pi}{4}-sin$ $\dfrac{\pi}{4}.cos2x=0,5$
$⇔$ $sin(2x-\dfrac{\pi}{4})=0,5$ $⇔$\(\left[ \begin{array}{l}2x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\)$⇔$ \(\left[ \begin{array}{l}2x=\dfrac{5\pi}{12}+k2\pi\\2x=\dfrac{13\pi}{12}+k2\pi\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{5\pi}{24}+k\pi\\x=\dfrac{13\pi}{24}+k\pi\end{array} \right.\) $(k∈Z)$
BẠN THAM KHẢO NHA!!!
