解下面三角方程:(1)cosx-2sinx=0 (2)sinx=cos3x (3)sin^2x-3cos^2x=sin2
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解下面三角方程:(1)cosx-2sinx=0 (2)sinx=cos3x (3)sin^2x-3cos^2x=sin2x
(4)3sin^2x+2sinx-1=0(5)2sinxcosx+sinx-cosx=1
(4)3sin^2x+2sinx-1=0(5)2sinxcosx+sinx-cosx=1
(1)cosx-2sinx=0
tanx=1/2
x=kπ +arctan(1/2),k∈Z
(2)sinx=cos3x
sin(2x-x)=cos(2x+x)
sin2xcosx-cos2xsinx=cos2xcosx-sin2xsinx
(sin2x-cos2x)(cosx+sinx)=0
tan2x=1 or tanx=-1
x=k(π/2)+π/8,or x=kπ+π/4,k∈Z
(3)sin^2x-3cos^2x=sin2x
-1-2cos2x=sin2x
sin2x+2cos2x=-1
sin2x=-1/√5,cos2x=-2/V5
2x=(2k+1)π+arcsin(1/√5),k∈Z
(4)3sin^2x+2sinx-1=0
sinx=1/3 or sinx=-1
x=kπ+(-1)^karcsin(1/3),or x=2kπ-π/2,k∈Z
(5)2sinxcosx+sinx-cosx=1
sinx-cosx=(sinx-cosx)²
sinx-cosx=0 or sinx-cosx=1
x=kπ+π/4,or x=kπ+π/2,or x=kπ+π,k∈Z
tanx=1/2
x=kπ +arctan(1/2),k∈Z
(2)sinx=cos3x
sin(2x-x)=cos(2x+x)
sin2xcosx-cos2xsinx=cos2xcosx-sin2xsinx
(sin2x-cos2x)(cosx+sinx)=0
tan2x=1 or tanx=-1
x=k(π/2)+π/8,or x=kπ+π/4,k∈Z
(3)sin^2x-3cos^2x=sin2x
-1-2cos2x=sin2x
sin2x+2cos2x=-1
sin2x=-1/√5,cos2x=-2/V5
2x=(2k+1)π+arcsin(1/√5),k∈Z
(4)3sin^2x+2sinx-1=0
sinx=1/3 or sinx=-1
x=kπ+(-1)^karcsin(1/3),or x=2kπ-π/2,k∈Z
(5)2sinxcosx+sinx-cosx=1
sinx-cosx=(sinx-cosx)²
sinx-cosx=0 or sinx-cosx=1
x=kπ+π/4,or x=kπ+π/2,or x=kπ+π,k∈Z
解下面三角方程:(1)cosx-2sinx=0 (2)sinx=cos3x (3)sin^2x-3cos^2x=sin2
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