\(sin\left(2x\right)=\frac{cos^4\left(x\right)}{2}-\frac{sin^4\left(x\right)}{2}\Leftrightarrow2sin\left(2x\right)=cos^4\left(x\right)-sin^4\left(x\right)\\ \Leftrightarrow2sin\left(2x\right)=\left(cos^2x\right)^2-\left(sin^2x\right)^2\)
\(\Leftrightarrow2sin2x=\left(\frac{1+cos2x}{2}\right)^2-\left(\frac{1-cos2x}{2}\right)^2\\ \Leftrightarrow2sin2x=\left(\frac{1+cos2x+1-cos2x}{2}\right)\left(\frac{1+cos2x-1+cos2x}{2}\right)\\ \Leftrightarrow2sin2x=\frac{4cos2x}{2}\)
\(\Leftrightarrow4sin2x=4cos2x\Leftrightarrow sin2x-cos2x=0\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\ \Leftrightarrow x-\frac{\pi}{4}=k\pi\\ \Leftrightarrow x=k\pi+\frac{\pi}{4}\)
