Решить уравнение. 1) sin3x=cos2x

2) 2sinx-3cosx=3

1

sin3x-sin (π/2-2x) = 0

2sin (x/2-π/4) cos (5x/2+π/4) = 0

sin (x/2-π/4) = 0⇒x/2-π/4=πn⇒x/2=π/4+πn⇒x=⇒/2+2⇒n, n∈z

cos (5x/2+π/4) = π/2+πk⇒5x/2=π/4+πk⇒x=π/10+2πk/5, k∈z

2

2sinx-3cosx-3=0

4sin (x/2) cos (x/2) - 3cos² (x/2) + 3sin² (x/2) - 3sin² (x/2) - 3cos² (x/2) = 0

4sin (x/2) cos (x/2) - 6cos² (x/2) = 0

2cos (x/2) * (2sin (x/2) - 3cos (x/2)) = 0

cos (x/2) = 0⇒x/2=π/2+πn⇒x=π+2πn, n∈z

2sin (x/2) - 3cos (x/2) = 0/cos (x/2)

2tg (x/2) - 3=0⇒tg (x/2) = 1,5⇒x/2=arctg1,5+πk⇒x=2arctg1,5+2πk, k∈z