Решить I sinx I=sinx+2cosx

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