Question

2sin*x-cos*x+sinxcosx-6sinx+3cosx=0,求2cos*x+sin2x除以1+tanx的值

Answer 1

2(sinx)^2-(cosx)^2+sinxcosx-6sinx+3cosx=0

(sinx+cosx)(2sinx-cosx)-3(2sinx-cosx)=0

(2sinx-cosx)(sinx+cosx-3)=0

因为sinx+cosx0

sinx*cosx=根号(1/5*4/5)=2/5

2(cosx)^2+sin2x

=2(cosx)^2+2sinxcosx

=8/5+4/5

=12/5

所以原式=(12/5)/(1+1/2)=8/5