Question

已知2sin^2(X)-cos^2(X)+sinXcosX-6sinX+3cosX=0，求2cosX(sinX+cosX)/1+tanX的值

已知2sin^2(X)-cos^2(X)+sinXcosX-6sinX+3cosX=0，求2cosX(sinX+cosX)/1+tanX的值

Answer 1

2sin^2(X)-cos^2(X)+sinXcosX-6sinX+3cosX=0
→→(2sinx-cosx)(sinx+cosx)-3(2sinx-cosx)=0
(2sinx-cosx)(sinx+cosx-3)=0
2sinx-cosx=0，sinx+cosx-3<0
2sinx = cosx
( 2sinx)^2=(cosx)^2,( sinx)^2+(cosx)^2=1;
(cosx)^2=4/5.
2cosX(sinX+cosX)/(1+tanX)
=2cosX(1+tanX)cosX/(1+tanX)
=2cos^2(X)=8/5