已知sin(x/2)-2cos(x/2)=0.

由题意知sin(x/2)=2cos(x/2)；即sin(x/2)/cos(x/2)=2=tan(x/2);
⑴二倍角公式;tan(2α)=2tanα/(1-tanα^2)故
tanx=2tan（x/2）/(1-tan(x/2)*tan(x/2))=-4/3;
⑵ 由和 差化积公式可求得
(cos2x)÷[√2 cos（π/4 + x）sinx]=(cosx+sinx)*(cosx-sinx))÷[√2( cosπ/4cosx-sinπ/4sinx)sinx]=
(cosx+sinx)/sinx=1/tanx+1=1/4

(1)tanx=2tan(x/2)/(1-（tan(x/2)）^2)
tanx=-4/3
(2）[(cosx)^2-(sinx)^2]/[(cosx-sinx)sinx]
=1+cotx=1/4

tan(x/2)=sin(x/2)/cos(x/2)=2 tanx=2tan(x/2)/1-tan(x/2)2=-4/3
(cos2x)÷[√2 cos（π/4 + x）sinx]=(cosx+sinx)*(cosx-sinx))÷[√2( cosπ/4cosx-sinπ/4sinx)sinx]=
(cosx+sinx)/sinx=1/tanx+1=1/4

(1) sin(x/2)-2cos(x/2)=0
tan(x/2)=2
tan(x/2)=tan(x-x/2)=[tanx-tan(x/2)]/[1+tanx*tan(x/2)]=2
[tanx-2]/[1+2tanx]=2
2+4tanx=tanx-2
tanx=-4/3