Question

решить I sinx I=sinx+2cosx

Answer 1

1)sinx<0⇒x∈(π+2πn;2π+2πn,n∈Z)
-sinx=sinx+2cosx
2sinx+2cosx=0
sinx+cosx=0/cosx≠0
tgx+1=0
tgx=-1
x=3π/4+2πn,n∈z
2)sinx≥0⇒x∈[2πn;π+2πn,n∈Z]
sinx=sinx+2cosx
2cosx=0
cosx=0
x=π/2+2πk,k∈Z