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

16(1-cos2x)²/4+8cos2x-7=04-8cos2x+4cos²2x+8cos2x-7=04cos²2x=34(1+cos4x)/2=31+cos4x=1,5cos4x=0,54x=+-π/3+2πk,k∈zx=+-π/12+πk/2,k∈z