$sin^4x+cos^4x=\dfrac{1}{2}$
$⇔(sin^2x)^2+2sin^2xcos^2x+(cos^2x)^2-2sin^2xcos^2x=\dfrac{1}{2}$
$⇔(sin^2x+cos^2x)^2-2sin^2xcos^2x=\dfrac{1}{2}$
$⇔1-2sin^2xcos^2x=\dfrac{1}{2}$
$⇔1-\dfrac{1}{2}(2sinxcosx)^2=\dfrac{1}{2}$
$⇔1-\dfrac{1}{2}(sin^22x)=\dfrac{1}{2}$
$⇔sin^22x-1=0$
$⇔$\(\left[ \begin{array}{l}sin2x=1=sin(\dfrac{\pi}{2})\\sin2x=-1=sin(\dfrac{-\pi}{2})\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k2\pi\\2x=\pi-\dfrac{\pi}{2}+k2\pi\end{array} \right.\) hay \(\left[ \begin{array}{l}2x=-\dfrac{\pi}{2}+k2\pi\\2x=\pi+\dfrac{\pi}{2}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{5\pi}{4}+k\pi\end{array} \right.\) hay \(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{7\pi}{4}+k\pi\end{array} \right.\)
