3) [tex]\dpi{120} \sqrt{2}\left ( sin\frac{5\pi}{12}+sin\frac{5\pi-24x}{12} \right )=1<=>\sqrt{2}sin\frac{5\pi-24x}{12}=1-\frac{\sqrt{3}+1}{2} <=>sin\frac{5\pi-24x}{12}=\frac{1-\sqrt{3}}{2\sqrt{2}} =>\frac{5\pi}{12}-2x=arcsin\frac{1-\sqrt{3}}{2\sqrt{2}}+k2\pi =>x=...[/tex]
hoac : [tex]\dpi{120} \frac{5\pi}{12}-2x=\pi-arcsin\frac{1-\sqrt{3}}{2\sqrt{2}}+k2\pi =>x=...[/tex]
4) [tex]\dpi{120} cos4x+cos2x+\sqrt{3}(1+sin2x)=\sqrt{3}(sin2x-cos2x)^2<=>cos4x+cos2x+\sqrt{3}(1+sin2x)=\sqrt{3}sin^22x-\sqrt{3}sin4x+\sqrt{3}cos^22x<=>cos4x+\sqrt{3}sin4x+cos2x+\sqrt{3}sin2x=0<=>2sin\left ( 4x+\frac{\pi}{6} \right )+2sin\left ( 2x+\frac{\pi}{6} \right )=0<=>sin\left ( 4x+\frac{\pi}{6} \right )=-sin\left ( 2x+\frac{\pi}{6} \right )<=>sin\left ( 4x+\frac{\pi}{6} \right )=sin\left ( -2x-\frac{\pi}{6} \right )[/tex]