Решить I sinx I=sinx+2cosx
1)sinx<0⇒x∈(π+2πn;2π+2πn,n∈Z)<br>-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