解
√3sin(2x-π/6)+1-cos(2x-π/6)
=2[√3/2sin(2x-π/6)-1/2cos(2x-π/6)]+1
=2[sin(2x-π/6)cosπ/6-sinπ/6cos(2x-π/6)]+1
=2sin[(2x-π/6)-π/6]+1
还有一个公式
asinx+bcosx=√a²+b²sin(x+θ)(tanθ=b/a)
解
√3sin(2x-π/6)+1-cos(2x-π/6)
=2[√3/2sin(2x-π/6)-1/2cos(2x-π/6)]+1
=2[sin(2x-π/6)cosπ/6-sinπ/6cos(2x-π/6)]+1
=2sin[(2x-π/6)-π/6]+1
还有一个公式
asinx+bcosx=√a²+b²sin(x+θ)(tanθ=b/a)