`~rai~`
\(\sin^4x+\cos^4x=\dfrac{3}{4}\\\Leftrightarrow (\sin^4x+2\sin^2x\cos^2x+\cos^4x)-2\sin^2x\cos^2x=\dfrac{3}{4}\\\Leftrightarrow (\sin^2x+\cos^2x)^2-\dfrac{1}{2}.4\sin^2x\cos^2x=\dfrac{3}{4}\\\Leftrightarrow 1-\dfrac{1}{2}\sin^22x=\dfrac{3}{4}\\\Leftrightarrow \dfrac{1}{2}.\dfrac{1-\cos4x}{2}=\dfrac{1}{4}\\\Leftrightarrow \dfrac{1-\cos4x}{4}=\dfrac{1}{4}\\\Leftrightarrow 1-\cos4x=1\\\Leftrightarrow \cos4x=0\\\Leftrightarrow 4x=\dfrac{\pi}{2}+k\pi\\\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{4}.(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{8}+k\dfrac{\pi}{4}\Big|k\in\mathbb{Z}\right\}.\)
