Ta có: \(\left(2\cdot cos2x-1\right)\left(\sin2x+cos2x\right)=1\)
=>\(2\cdot cos2x\cdot\sin2x+2\cdot cos^22x-\sin2x-cos2x=1\)
=>\(\sin4x+1+cos4x-\sin2x-cos2x=1\)
=>\(\sin4x+cos4x-\sin2x-cos2x=0\)
=>\(\left(\sin4x-\sin2x\right)+\left(cos4x-cos2x\right)=0\)
=>\(2\cdot cos\left(\frac{4x+2x}{2}\right)\cdot\sin\left(\frac{4x-2x}{2}\right)+\left(-2\right)\cdot\sin\left(\frac{4x+2x}{2}\right)\cdot\sin\left(\frac{4x-2x}{2}\right)=0\)
=>\(2\cdot cos3x\cdot\sin x-2\cdot\sin3x\cdot\sin x=0\)
=>\(\sin x\cdot\left(cos3x-\sin3x\right)=0\)
TH1: sin x=0
=>\(x=\frac{\pi}{2}+k2\pi\)
TH2: cos3x-sin 3x=0
=>sin 3x-cos3x=0
=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=0\)
=>\(\sin\left(3x-\frac{\pi}{4}\right)=0\)
=>\(3x-\frac{\pi}{4}=k\pi\)
=>\(3x=\frac{\pi}{4}+k\pi\)
=>\(x=\frac{\pi}{12}+\frac{k\pi}{3}\)
