Question

2 cos4x + 2 cos2x = 4 sin^2 x -1

Answer 1

2(2сos²2x-1)+2cos2x=4*(1-cos2x)/2 -1
4os²2x-2+2cos2x-2+2cos2x+1=0
4cos²2x+4cos2x-3=0
cos2x=a
4a²+4a-3=0
D=16+48=64
a1=(-4-8)/8=-1,5⇒cos2x=-1,5 нет решения
a2=(-4+8)/8=1/2⇒cos2x=1/2⇒2x=+-π/3+2πn⇒x=+-π/6+πn