Sin^2 x=1-cos x решите уравнение

Sin²x=1-cosx

1-cos²x=1-cosx

-cos²x=-cosx

cosx-cos²x=0

cosx (1-cosx) = 0

a) cosx=0, x=π/2+kπ, k∈Z

b) 1-cosx=0, cosx=1, x=2kπ, k∈Z

Помойму так 1) Sin (x-1) = cos (x+2)

Sin (x-1) - cos (x+2) = 0

Sin (x-1) - sin (π/2 - (x+2)) = 0

2sin (π/4-3/2) * cos (x+1/2-π/4) = 0

cos (x+1/2-π/4) = 0

x+1/2-π/4=π/2+πn

x=π/2-1/2+π/4+πn

x=3π/2-1/2+πn

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=π/2+πn

x1=π/10+πn/5

б) cosx=0

x2=π/2+πn