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已知函数 F(x)=2sin(2x+ π 6 ),x∈R .(...

(1)∵f(x)=2sin(2x+ π 6 ),∴其最小正周期T= 2π 2 =π;∴由2kπ- π 2 ≤2x+ π 6 ≤2kπ+ π 2 得kπ- π 3 ≤x≤kπ+ π 6 (k∈Z),∴函数的增区间为[kπ- π 3 ,kπ+ π 6 ](k∈Z),(2)∵x∈( π 4 , 3π 4 ],∴2x+ π 6 ∈( 2π 3 , 5π 3 ],∴-1≤sin(2...

∵函数f(x)=2sin(ωx+φ)对任意x都有 f( π 6 +x)=f( π 6 -x) ∴函数图象的对称轴是 x= π 6 ,∴ f( π 6 ) 取最大值或者是最小值∵函数的最大值是2,最小值是-2∴ f( π 6 ) 等于-2或2故选B.

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

π/4≤x≤3π/4, π/2≤2x≤3π/2, 2π/3≤2x+π/6≤5π/3, -2≤2sin(2x+π/6)≤√3, -2≤f(x)≤√3. a>0 b≤F(x)≤√3a+2a+b b=-√3, √3a+2a+b=(√3)-1, a=(2(√3)-1)/(√3+2)>0, ab=(√3-6)/(√3+2)。 a

(1)根据函数 f(x)=2sin(2x- π 6 ),x∈R ,可得函数的最小正周期为 2π 2 =π,f(0)=2sin(- π 6 )=2×(- 1 2 )=-1.(2)令 2kπ- π 2 ≤2x- π 6 ≤2kπ+ π 2 ,k∈z,求得 kπ- π 3 ≤x≤kπ+ π 3 ,故函数的增区间为[kπ- π 3 ,kπ+ π 3 ],k∈z.(...

(1)∵函数f(x)=sin(2x-π6),∴f(π4)=sinπ3=32.(2)当且仅当2x-π6=2kπ+π2,k∈z时,即x=kπ+π3时,该函数取得最大值1,所以该函数取得最大值时自变量的取值集合为{x|x=kπ+π3,k∈z}.(3)由f(α+π3)=35,求得cos2α=35=1-2sin2α,∴sinα=±55...

(I)∵函数f(x)=sin(7π6?2x)+2cos2x?1=sin7π6cos2x-cos7π6sin2x+cos2x=32sin2x+12cos2x=sin(2x+π6).故函数f(x)的周期为T=2π2=π.再令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,k∈z,故单调递增区间为[kπ-π3,kπ+π6],k∈z.(II)...

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

(1) f(x)=sin(2(x-π/12)) x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6] 单调递减区间是2(x-π/12)∈[π/2,3π/2] 即x-π/12∈[π/4,3π/4] 则x∈[π/3,5π/6] (2)x∈[-π/12,π/2]时, 2(x-π/12)∈[-π/3,5π/6] 而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),...

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