16sinx^4+8cos2x-7=0. Тригономеирия, срочно

16(1-cos2x)²/4+8cos2x-7=0 4-8cos2x+4cos²2x+8cos2x-7=0 4cos²2x=3 4(1+cos4x)/2=3 1+cos4x=1,5 cos4x=0,5 4x=+-π/3+2πk,k∈z x=+-π/12+πk/2,k∈z