求方程cosx=sin2x在0到2π的解
问题描述：
求方程cosx=sin2x在0到2π的解
答
cosx=sin2x
sin2x - cosx = 0
2sinxcosx - cosx =0
(2sinx-1)cosx=0
cosx=0或sinx=1/2
在0到2π的解
x=π/2或x=3π/2或x=π/6或x=5π/6
答
cosx=sin2x
cosx=2sinxcosx
(1)若cosx=0
则x=π/2或x=3π/2
(2)若cosx≠0
则1=2sinx
即sinx=1/2
所以x=π/6或x=5π/6
综上，x=π/2或x=3π/2或x=π/6 或x=5π/6
答
cosx=2sinxcosx
(2sinx-1)cosx=0
sinx=1/2,cosx=0
0