sin^2 x=1-cos x решите уравнение

Sinx=1-cosx
1-cosx=1-cosx
-cosx=-cosx
cosx-cosx=0
cosx(1-cosx)=0
a)cosx=0, x=/2+k,kZ
b)1-cosx=0, cosx=1, x=2k,kZ

 

Помойму так 1) Sin (x-1) = cos (x+2)
Sin (x-1) - cos (x+2)=0
Sin (x-1) - sin(/2- (x+2))=0
2sin(/4-3/2)*cos(x+1/2-/4)=0
cos(x+1/2-/4)=0
x+1/2-/4=/2+n
x=/2-1/2+/4+n
x=3/2-1/2+n

2) Sin x Sin 5x = Cos 4x;
1/2(cos(-4x)-cos6x)-cos4x=0
cos4x-cos6x-2cos4x=0
cos6x+cos4x=0
2cos5x*cosx=0
a) cos5x=0
5x=/2+n
x1=/10+n/5
б) cosx=0
x2=/2+n