Question

Решить уравнение 16sin^4x+8cos2x-7=0

Answer 1

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