Question

giải phương trình : sin² 2x=\(\dfrac{1}{2}\)

Answer 1

Ta có: \(sin^22x=\dfrac{1}{2}\Leftrightarrow\)\(\left[{}\begin{matrix}sin2x=\dfrac{1}{\sqrt{2}}\\sin2x=\dfrac{-1}{\sqrt{2}}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left[{}\begin{matrix}2x=\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\\left[{}\begin{matrix}2x=\dfrac{-\pi}{4}+k2\pi\\2x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{8}+k\pi\\x=\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{-\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

Answer 2

Refer:

Answer 3

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos4x=\dfrac{1}{2}\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow4x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)