Question

Решите уравнение
sinx*cosx+2sin^2x-cos^2x=0

Answer 1

sinx*cosx+2sin²x-cos²x=0 |:cos²x
cos²x≠0
cosx≠0
x≠п/2+пn,n€z

tgx+2tg²x-1=0
tgx=t
2t²+t-1=0
D=1-4*2*(-1)=9
t1=-1+3/4=1/2
t2=-4/4=-1
tgx=1/2
x=arctg1/2+пn,n€z
tgx=-1
x=-п/4+пn,n€z