Решите уравнение: а) sin(п+x)=cos(-п/3) б) sin2x=sin(x+п/2), при x=(5/6п; 3п)

A
-sinx=1/2
sinx=-1/2
x=(-1)^(n+1)*π/6+πn,n∈z
Б
2sinxcosx-cosx=0
cosx(2sinx-1)=0

cosx=0⇒x=π/2+πk,k∈z
5π/6<π/2+πk<3π
5<3+6k<18
2<6k<15
1/3<k<2 1/2
k=1⇒x=π/2+π=3π/2
k=2⇒π/2+2π=5π/2

sinx=1/2⇒x=π/6+2πn U x=5π/6+2πn
5π/6<π/6+2πn<3π  U  5π/6<5π/6+2πn<3π
5<1+12n<18  U  5<5+12n<18
4<12n<17  U  0<12n<13
1/3<n1 5/12   U  0<n<1 1/12
n=1⇒x=π/6+2π=13π/6  U  n=1⇒x=5π/6+2π=17π/6