解下面三角方程：(1)cosx－2sinx＝0 (2)sinx＝cos3x (3)sin＾2x－3cos＾2x＝sin2x

(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