Question

решить уравнение
1. 2sin^2x-sinxcosx=cos^2x

Answer 1

2sin^2x-sinxcosx=cos^2x
2sin^2x-sinxcosx-cos^2x=0 |÷cos2^x
2tg^2x-tgx-1=0
tgx=t
2t^2-t-1=0
D=1-4×2×(-1)=1+8=9
x1=1-3/4=-2/4=-1/2
x2=1+3/4=1

tgx=-1/2
x=arctg (-1/2)+pi×n

tgx=1
x=pi/4+pi×n ,n $Z