f x sinx 根号3cosx

f(x)=sinx 根号3cosx=根号3 *1/2 * sin(2 x)通解=x = pi* n, n为整数

f(x)'=√3cosx-(-sinx)-1 =√3cosx+sinx-1 =2sin(x+π/3)-1.单调增区间为：2kπ-π/2<=x+π/3<=2kπ+π/2 2kπ-5π/6<=x<=2kπ+π/6 单调减区间为：2kπ+π/2<=x+π/3<=2kπ+3π/2 2kπ+π/6<=x<=2kπ+7π/6 ...

首先利用公式cos2x=cos^2(x)-sin^2(x)=1-2sin^2(x),得sin^2(x)=(1-cos2x)将其与sinx*cosx=(1/2)sin2x带入原式中得：f(x)=(√3/2)sin2x-(1/2)cos2x-1/2=sin2x*cos(π/6)-cos2x*sin(π/6)-1/2=sin(2x-π/6)-1/2 所以最小正周期T=2π/2=π 由于正弦函数的...

f(x)=sinx+√3cosx=2sin(x+π/3) x属于（0，π/2）x+π/3属于（π/3,5π/6）当x+π/3=π/2时，f（x）有最大值 故x=π/6 （2）画出2sint在（π/3,5π/6）上的图像 方程f(x)-a=0 有两个实数根 故2sinπ/3<a<2 得到√3<a<2 ...

解：f(x)=√3sinxcosx-sin^2x.f(x)=(√3/2)sin2x-(1-cos2x)/2.=sin(2x+π/6)-1/2.(1) T=2π/2=π, ---所求函数f(x)的 最小正周期；(2)∵|sin(2x+π/6)|≤1.∴令2x+π/6=π/2.2x=π/2-π/6, 则x=π/6, 即当x=π/6时，sin(2x+π/3)=1,此时，函数...

因为原式可以变成f{x}=sinx+根号3cosx=2（1/2*sinx+(根号3)/2*cosx）=2（cos60*sinx+sin60*cosx）=2sin(x+π/3)所以最小正周期T=2π/1=2π

f(x)=√3cosxsinx+(cosx-1)(cosx+1)+1/2 =√3cosxsinx+cos²x-1+1/2 =√3/2sin2x+(1+cos2x)/2-1/2 =√3/2sin2x+1/2cos2x =sin(2x+π/6)T=2π/2=π 周期π

解：f(x)=sinx*(√3cosx-sinx).=√3sinxcosx-sin^2x,=(√3/2)2sinxcosx-(1-cos2x)/2.=(√3/2)sin2x+(1/2)cos2x-1/2.=sin(2x+π/6)-1/2.f(x)的最小正周期T=2π/2=π.∵x∈(0,2π/3). ∴(2x+π/6)∈(π/6, 3π/2).f(x)=sin(2x+π/6)＞-1-1/2...

f(x)=cosxsinx-√3cos²x+1 =(sin2x)/2-√3(cos2x+1)/2+1 =(1/2)×sin2x-(√3/2)×cos2x-√3/2+1 =sin(2x-π/3)＋(1-√3/2)类似的题目都可以这样化简 asinx+bcosx =√(a²+b²)sin(x＋α)其中，tanα=b/a ...

f(x)=2sin(x-π/6)(1)由-π/2+2kπ≤x-π/6≤π/2+2kπ得：-π/3+2kπ≤x≤2π/3+2kπ 所以单调增区间为：【-π/3+2kπ，2π/3+2kπ】2）你的题目条件，b小于3，不正确，因为A=60度，2b=3c纯属于两个条件，而解三角形必须三个独立条件，而 b<3,不能定位三角形，因此...