Question

Решить уравнение

1) sin 4x=-√2/2

2) cos(x/2-π/8)=0

3) cos3x+cosx=0

 

Answer 1

(1) 4х = (-1)^n * arcsin(-√2/2) + πn, n€Z

4х = (-1)^(n+1) * π/4 + πn, n€Z

х = (-1)^(n+1) * π/16 + πn/4, n€Z

 

(2) (x/2-π/8)=π/2+πn, n€Z

x/2 = π/8 + π/2+πn, n€Z

x/2 = 5π/8+πn, n€Z

x = 5π/4+2πn, n€Z

 

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

cos2x*sinx=0

cos2x=0 или sinx=0 (запись можно в виде совокупности)

1)2х=π/2+πn, n€Z

х=π/4+πn/2, n€Z

2)х=πk, k€Z