Question

1) tg 2x=ctg x

 

2) cos x=cos 2x-cos 3x

Answer 1

2)(cosx+co3x)-cos2x-0
2cos2xcosx-cos2x-0
cos2x(2cosx-1)=0
cos2x=0, 2x=π/2+πn,  x=π/4+πn/2
cosx=1/2,  x=+ - π/3+2πn
1)2tgx/(1-tg²x)=1/tgx
2tg²x=1-tg²x
3tg²x=1
tg²x=1/3
tgx= + - 1/√3
x=+ - arctgx1/√3+πn