Question

Помогите решить 1) cos+sin(pi/2+x)=1 2) tg^2x+tgx=0 3) 5-4sin^2x=4cosx

Answer 1

1) cosx+sin(pi/2+x)=1
cosx+cosx=1
2cosx=1
cosx=1/2
x=+-π/3+2πn,n∈z
2) tg^2x+tgx=0
tgx(tgx+1)=0
tgx=0⇒x=πn,n∈z
tgx=-1⇒x=-π/4+πk,k∈z
3) 5-4sin^2x=4cosx
5-4+4cos²x-4cosx=0
4cos²x-4cosx+1=0
(2cosx-1)²=0
2cosx=1
cosx=1/2
x=+-π/3+2πn,n∈z