y=sin(兀/6-2x)(x∈[0,兀])的单调递增区间是?
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y=Sin(-2-6/兀)x∈(0,兀】为增函数的区间
y=sin(兀/6-2x)(x∈[0,兀])的单调递增区间是?
.函数 y=sin(兀/2+x)cos(兀/6-x)的最大值
y=2sin(2x+π/6)x∈[0,π/2],求值域
5cos(2x-y)+7cosy=0 ,tan(x-y)tanx=?y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域
y=sin(x-π/6) x∈【0,π】函数y=sin(x-π/6),x∈【0,π】的值域是
已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos(x+y)+cos(x-y)】的值
求函数y=sin(x+1/3兀)sin(x+1/2兀)的周期
已知函数f(x)=[2sin(x-π/6)+√3sin x]cos x+sin^2x,x∈R
y=sin(兀/3-1/2x),x∈[-2兀,2兀]增区间?
y=sin(2x+a)(0
y=sin(2x+a)(0
x=3分之兀是不是y=sin(2x-6分之兀)的对称轴
函数y=sin(x+兀/2)cos(x+兀/6)的递减区间是
函数y=2sin(兀x/6-兀/3)(0
函数y=2sin(兀x/6-兀/3)(0
y=1-x^2 x>=0 ,y=sin|x|/x x
函数y=2sin(pai/6-2x)(x∈0,pai)