方程2sin(x π 3) a-1=0

由根与系数的关系sina+cosa=2/3sinacosa=a/3sina^2+cosa^2=(sina+cosa)^2-2sinacosa=4/9-2a/3=1a=-5/6

tan(A+B)=-12sin^2(A+B)-3sin(A+B)cos(A+B)+cos^2(A+B)=[2sin^2(A+B)-3sin(A+B)cos(A+B)+cos^2(A+B)]/[sin^

∵-π/6≤x≤13π/12∴0≤2x+π/3≤5π/2sin(2x+π/3)=a+1令t=2x+π/3∈[0,5π/2]sint=a+1（画出sint在[0,5π/2]的函数图像）a+1=1或-1≤

2sin(x+π/3)+2a-1=0,就是2a=1-2sin(x+π/3)在[0,π]上的取值范围x∈[0,π],(x+π/3)∈[π/3,4π/3],2sin(x+π/3)∈[－√3,2],于是1-

1解因为函数F(X)=sin(x+π/3)+asin(x-π/6)（a>0）的一条对称轴方程为:x=π/2,0和π关于π/2对称所以F(0)=F(π)sin(π/3)+asin(-π/6)=sin(4

方程化简为：2sin（x+π4）=a即sin（x+π4）=2a2      若没有解集，那么2a2>1或2a2<-1解得&nbs

解题思路：同学你好，你给的题目，条件不明确。我解答时，添加的条件是【若该方程在区间[-π/6,13π/12]上面有解】，若题目还有其他情况，请添加讨论.解题过程：

1.原式=(sin^2a/(sina-cosa))+(cos^2a/(cosa-sina))=sina+cosa=(根3+1)/22.sina+cosa=(根3+1)/2sin^2a+cos^2a=1

x^2-ax+a=0韦达定理得:sina+cosa=a,sinacosa=a所以(sina)^3+(cosa)^3=(sina+cosa)[(sina)^2-sinacosa+(cosa)^2]=a*

sin^2(a+b)+6sin(a+b)cos(a+b)+5cos^2(a+b)=sin^2(a+b)+6sin(a+b)cos(a+b)+5cos^2(a+b)/sin^2(a+b)+cos^2(a

f(x)=2sin(2/3πx+π/6).容易求对称轴即2πx/3+π/6=π/2+2kπf(x)=a(1

F(x)=sin(3x-45°)=a3x-45°=arcsina3x=arcsina+45°0

f(x)=sin(x+π/3)-asin(x-π/6)=sin(x+π/3)-asin[(x+π/3)-π/2]=sin(x+π/3)+acos(x+π/3),因为函数图像的一条对称轴方程为x=π/2

cos(x-π/4)=sin2x+acosxcos(π/4)+sinxsin(π/4)=2sinxcosx+a(根号2/2)*cosx+(根号2/2)*sinx=2sinxcosx+a(根号2/2)(

本题就相当于求方程y=sin(x)=a(0＜a＜1）在【-π/4,23/4π】内的所有实数根之和；你画下y=sinx和y=a(0＜a＜1）的图像可知道,就中间两个峰被截到；因为每个峰是对称的,不管怎么

由韦达定理可以得到tan(a)+tan(b)=3,tan(a)*tan(b)=-3,所以tan(a+b)=3/4.也就是说sin(a+b)/cos(a+b)=3/4.因此sin(a+b)=3/5,co

解此方程,得tana=根号2/2+1tanb=-根号2/2+1原式=cos^(a+b)·又∵1/cos^(a+b)=tan^(a+b)+1且tan(a+b)=(tana+tanb)/(1-tana·t

答：0

f(x)=(2cos^2a/2-sina-1)/(sin(a+π/3)sin(a-π/3)=4(cosa-sina)/(sin^2a-3cos^2a)求啥?

sin2x+sinx=02sinxcosx+sinx=0sinx(2cosx+1)=0①党sinx=0x=kπ②党(2cosx+1)=0cosx=-1/2x=kπ/2+π/3