已知关于X的方程X(cosA-3/5)+sinA=4/5-sinB- XcosB的解集是R,求cos(A-B)的值
x(cosA-3/5)+sinA=4/5-sinB-xcosB
x(cosA+cosB-3/5)=4/5-sinB-sinA
由于:解集X属于R
则有:cosA+cosB-3/5=0
且:4/5-sinB-sinA=0
则：cosA+cosB=3/5-----(1)
sinA+sinB=4/5-----(2)
(1)^2+(2)^2
[cos^2(A)+cos^2(B)+2cosAcosB]+[sin^2(A)+sin^2(B)+2sinAsinB]=(3/5)^2+(4/5)^2
[cos^2(A)+sin^2(A)]+[cos^2(B)+sin^2(B)]
+2(cosAcosB+sinAsinB)=1
即：2+2cos(A-B)=1
则：cos(A-B)=-1/2