Решить уравнение: cos6x-cos3x=0
cos6x=2cos²3x-1
cos6x-cos3x=0
2cos²3x-cos3x-1=0
cos3x=-1/2 или cos3x=1
3x=±arccos(-1/2)+2πn, 3x=2πk
3x=±(π-π/3)+2πn, x=2πk/3, k∈Z
x=±2π/9+2πn/3, n∈Z
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cos6x=2cos²3x-1
cos6x-cos3x=0
2cos²3x-cos3x-1=0
cos3x=-1/2 или cos3x=1
3x=±arccos(-1/2)+2πn, 3x=2πk
3x=±(π-π/3)+2πn, x=2πk/3, k∈Z
x=±2π/9+2πn/3, n∈Z