Решить уравнение
1) sin 4x=-√2/2
2) cos(x/2-π/8)=0
3) cos3x+cosx=0
(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
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Verified answer
(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