\(cos2x-sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left(cos2x-sin2x\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left(cos^2x+sin^2x\right)+2sin2xcos2x=\dfrac{1}{4}\)
\(\Leftrightarrow1+sin4x=\dfrac{1}{4}\)
\(\Leftrightarrow sin4x=-\dfrac{3}{4}\)
Lời giải:
\(\cos 2x=\sin 2x+\frac{1}{2}\Rightarrow (\sin 2x+\frac{1}{2})^2=\cos ^22x=1-\sin ^22x\)
\(\Rightarrow \sin 2x=\frac{-1\pm \sqrt{7}}{4}\)
\(\Rightarrow \cos 2x=\frac{1+\sqrt{7}}{4}\) nếu $\sin 2x=\frac{-1+\sqrt{7}}{4}$ và $\cos 2x=\frac{1-\sqrt{7}}{4}$ nếu $\sin 2x=\frac{-1-\sqrt{7}}{4}$
Do đó:
$\sin 4x=2\sin 2x\cos 2x=\frac{3}{4}$
cos2x-sin2x=1/2
<=>(cos2x-sin2x) ^2=1/4
<=>(cos^2+sin^2)+2sin2x2cos2x=1/4
<=>1+sin4x=1/4
<=>sin4x=-3/4
