Question

Giải phương trình \(sin^2\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

Answer 1

\(\sin^2\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\Leftrightarrow\dfrac{1-\cos\left(4x-\dfrac{2\pi}{3}\right)}{2}=\dfrac{1}{2}\Leftrightarrow\cos\left(4x-\dfrac{2\pi}{3}\right)=0\Leftrightarrow4x-\dfrac{2\pi}{3}=\dfrac{\pi}{2}+k\pi\Leftrightarrow x=\dfrac{7\pi}{24}+\dfrac{k\pi}{4}\)