答:
f(x)=2√3cos²x+sin2x-√3+1
f(x)=√3(cos2x+1)+sin2x-√3+1
f(x)=√3cos2x+sin2x+1
f(x)=2*[(1/2)sin2x+(√3/2)cos2x]+1
f(x)=2sin(2x+π/3)+1
1)
f(x)的最小正周期T=2π/2=π
单调递增区间满足:
2kπ-π/2
答:
f(x)=2√3cos²x+sin2x-√3+1
f(x)=√3(cos2x+1)+sin2x-√3+1
f(x)=√3cos2x+sin2x+1
f(x)=2*[(1/2)sin2x+(√3/2)cos2x]+1
f(x)=2sin(2x+π/3)+1
1)
f(x)的最小正周期T=2π/2=π
单调递增区间满足:
2kπ-π/2