Question

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

Answer 1

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