cos4x=sin2x+cos2x
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/06/04 06:08:43
cos4x=sin2x+cos2x
cos4x=cos²2x-sin²2x=cos2x+sin2x
(cos2x+sin2x)(cos2x-sin2x)-(cos2x+sin2x)=0
(cos2x+sin2x)(cos2x-sin2x-1)=0
cos2x+sin2x=0,cos2x-sin2x=1
cos2x+sin2x=0
√2sin(2x+π/4)=0
2x+π/4=kπ
x=kπ/2-π/8
cos2x-sin2x=1
sin2x-cos2x=-1
√2sin(2x-π/4)=-1
sin(1x-π/4)=-√2/2
2x-π/4=2kπ-3π/4,2x-π/4=2kπ-π/4
x=kπ-π/4,x=kπ
所以
x=kπ/2-π/8,x=kπ-π/4,x=kπ
(cos2x+sin2x)(cos2x-sin2x)-(cos2x+sin2x)=0
(cos2x+sin2x)(cos2x-sin2x-1)=0
cos2x+sin2x=0,cos2x-sin2x=1
cos2x+sin2x=0
√2sin(2x+π/4)=0
2x+π/4=kπ
x=kπ/2-π/8
cos2x-sin2x=1
sin2x-cos2x=-1
√2sin(2x-π/4)=-1
sin(1x-π/4)=-√2/2
2x-π/4=2kπ-3π/4,2x-π/4=2kπ-π/4
x=kπ-π/4,x=kπ
所以
x=kπ/2-π/8,x=kπ-π/4,x=kπ
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