急!高一数学题设函数f(x)＝2cos^2x＋根号3sin2x-2
急!高一数学题设函数f(x)＝2cos^2x＋根号3sin2x-2
1.求f(π/3)的值 2.求函数最小正周期及函数的单调递增区间
1.求f(π/3)的值 2.求函数最小正周期及函数的单调递增区间
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f(x)＝2cos^2x＋根号3sin2x-2
=cos2x+1+√3sin2x-2
=2sin(2x+π/6)-1
f(π/3)=2sin(5π/6)-1=1-1=0
最小正周期为 2π/2=π
单调增区间：
2x+π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/3,kπ-π/6]
所以 单调增区间为 ∈[kπ-π/3,kπ-π/6] k∈z
=cos2x+1+√3sin2x-2
=2sin(2x+π/6)-1
f(π/3)=2sin(5π/6)-1=1-1=0
最小正周期为 2π/2=π
单调增区间：
2x+π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/3,kπ-π/6]
所以 单调增区间为 ∈[kπ-π/3,kπ-π/6] k∈z
1年前
1
共回答了1627个问题 举报
f(x) = 2(cosx)^2 +√3sin2x-2
= cos2x + √3sin2x -1
= 2(sin(2x+π/6)) -1
f(π/3) = 2sin(5π/6) -1
= 2(1/2) -1
= -3/4
最小正周期 = π
单调递增区间
2nπ -π/2 <= 2x+π/6 <= 2nπ +π/2
nπ-4π/3 <=x <= nπ +π/6
= cos2x + √3sin2x -1
= 2(sin(2x+π/6)) -1
f(π/3) = 2sin(5π/6) -1
= 2(1/2) -1
= -3/4
最小正周期 = π
单调递增区间
2nπ -π/2 <= 2x+π/6 <= 2nπ +π/2
nπ-4π/3 <=x <= nπ +π/6
1年前
2
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