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cos4x=sin2x+cos2x
人气:297 ℃ 时间:2020-05-21 11:13:33
解答
cos4x=cos²2x-sin²2x=cos2x+sin2x(cos2x+sin2x)(cos2x-sin2x)-(cos2x+sin2x)=0(cos2x+sin2x)(cos2x-sin2x-1)=0cos2x+sin2x=0,cos2x-sin2x=1cos2x+sin2x=0√2sin(2x+π/4)=02x+π/4=kπx=kπ/2-π/8cos2x-s...
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