已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx求:(1)求f(x)的最小正周期 (2)设x∈[-π/3,π/3] ,求f(x)的知遇和单调递增区间
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![已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx求:(1)求f(x)的最小正周期 (2)设x∈[-π/3,π/3] ,求f(x)的知遇和单调递增区间](/uploads/image/z/1618477-61-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E2%88%9A3%28sin%26%23178%3Bx-cos%26%23178%3Bx%29-2sinxcosx%E6%B1%82%3A%281%29%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F+%282%29%E8%AE%BEx%E2%88%88%5B-%CF%80%2F3%2C%CF%80%2F3%5D+%2C%E6%B1%82f%28x%29%E7%9A%84%E7%9F%A5%E9%81%87%E5%92%8C%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx求:(1)求f(x)的最小正周期 (2)设x∈[-π/3,π/3] ,求f(x)的知遇和单调递增区间
已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx
求:(1)求f(x)的最小正周期 (2)设x∈[-π/3,π/3] ,求f(x)的知遇和单调递增区间
已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx求:(1)求f(x)的最小正周期 (2)设x∈[-π/3,π/3] ,求f(x)的知遇和单调递增区间
f(x)=√3(sin^2x-cos^2x)-2sinxcosx
=-√3cos2x-sin2x
=-2sin(2x+π/3)
1.求最小正周期T=π
2.设x∈[-π/3,π/3],求函数的值域和单调区间
-π/2
解f(x)=√3(sin^2x-cos^2x)-2sinxcosx
=-√3cos2x-sin2x
=-2(√3/2cos2x+1/2sin2x)
=-2sin(2x+π/3)
知1函数的T=2π/2=π,
2由x∈[-π/3,π/3]
知2x∈[-2π/3,2π/3]
即2x+π/3∈[-π/3,π]
即sin(2x+π/3)...
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解f(x)=√3(sin^2x-cos^2x)-2sinxcosx
=-√3cos2x-sin2x
=-2(√3/2cos2x+1/2sin2x)
=-2sin(2x+π/3)
知1函数的T=2π/2=π,
2由x∈[-π/3,π/3]
知2x∈[-2π/3,2π/3]
即2x+π/3∈[-π/3,π]
即sin(2x+π/3)∈[-√3/2,1]
即-2sin(2x+π/3)∈[-2,√3]
即y∈[-2,√3]
故函数的值域[-2,√3]
由2x+π/3∈[-π/3,π]
知当2x+π/3∈[-π/3,π/2]
即当2x∈[-2π/3,π/6]
即x∈[-π/3,π/12] 时,y=√3(sin²x-cos²x)-2sinxcosx是增函数
故函数的增区间为[-π/3,π/12] 。
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f(x)=√3(sin²x-cos²x)-2sinxcosx=-√3cos2x-sin2x=-2sin(2x+π/3)
f(x)最小正周期 π
f(x)在[-2,2]
单增 2kπ+π/2≤2x+π/3≤ 2kπ+3π/2
kπ+π/12≤x≤ kπ+7π/12
f(x)=√3(sin²x-cos²x)-2sinxcosx
=-√3cos2x-sin2x
=-2(√3/2cos2x-1/2sin2x)
=-2cos(2x+π/6)
=2cos[π+(2x-5π/6)]
=2cos(2x-5π/6)]
T=2π/2=π
x∈[-π/3,π/3]
2x-5π/6∈[-3π/2,-π/6]
值域为[ -√3,0]