Question

Решите уравнения: 1. cos²x-sin²x= -1 2. 2sinxcox=0 3. 1+cos6x=cos3x

Answer 1

1) cos(x)^2 - sin(x)^2 = -1
1-sin(x)^2 - sin(x)^2 = -1
2sin(x)^2 = 2
x = (-1)^n * p/2 + n*p

2) 2cos(x)*sin(x) = 0, значит либо sin(x) = 0, либо cos(x) = 0
sin(x) = 0  x = 0 + n*p
cos(x) = 0  x = p/2 + n*p

3) 1 + cos(6x) = cos(3x)
1 + cos(3x)^2 - sin(3x)^2 - cos(3x) = 0
2cos(3x)^2 - cos(3x) = 0
2cos(3x)*(cos(3x)-1/2) = 0
cos(3x) = 0  x = p/2 + n*p
cos(3x) = 1/2  x = p/9 + n*p/3