=>(1+cos2x)/2-sin2x+7(1-cos2x)/2=5
=>1/2+1/2cos2x-sin2x+7/2-7/2cos2x=5
=>-sin2x-3cos2x=5-7/2-1/2=5-4=1
=>sin2x+3cos2x=-1
=>\(\dfrac{1}{\sqrt{10}}\cdot sin2x+\dfrac{3}{\sqrt{10}}\cdot cos2x=-\dfrac{1}{\sqrt{10}}\)
=>sin(2x+a)=-cosa=cos(pi-a)
=>sin(2x+a)=sin(pi/2-pi+a)=sin(a-pi/2)
=>2x+a=a-pi/2+k2pi hoặc 2x+a=3/2pi-a+k2pi
=>x=-pi/4+kpi hoặc x=3/4pi-a+kpi