Question

4cos4x+6sin^2x+5cos2x=0

Answer 1

8cos²2×-4 6-6cos²2x 5cos2x=0
2cos²2x 5cos2x 2=0
cos2x=a
2a² 5a 2=0
D=25-16=9
a1=(-5-3)/4=-2⇒cos2x=-2∉[-1;1]
a2=(-5 3)/4=-1/2⇒cos2x=-1/2⇒2x= -2π/3 2πn⇒x= -π/3 πn