a) $5(sinx-cosx)^2=cos4x+2$
$↔ 5(1-sin2x)=1-2sin^22x+2$
$↔ 5-5sin2x=1-2sin^22x+2$
$↔ 2sin^22x-5sin2x+2=0$
$↔ \left[ \begin{array}{l}sin2x=2\\sin2x=\dfrac{1}{2}\end{array} \right.$
Vì $sin2x∈[-1;1]$ nên loại $sin2x=2$
$sin2x=\dfrac{1}{2}$
$↔ sin2x=sin\dfrac{\pi}{6}$
$↔ \left[ \begin{array}{l}2x=\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\end{array} \right.$ $(k∈Z)$
b) $2sin^2xcos^2x-sin2x+\dfrac{1}{2}=0$
$↔ 2\Bigg(\dfrac{1}{2}sin2x\Bigg)^2-sin2x+\dfrac{1}{2}=0$
$↔ \dfrac{1}{2}sin^22x-sin2x+\dfrac{1}{2}=0$
$↔ sin^22x-2sin2x+1=0$
$↔ (sin2x-1)^2=0$
$↔ sin2x-1=0$
$↔ sin2x=1$
$↔ 2x=\dfrac{\pi}{2}+k2\pi$
$↔ x=\dfrac{\pi}{4}+k\pi$ $(k∈Z)$
