Question

1) cos²x cos 2x = sin x; ​

Answer 1

1-cos2x=sinx

1-(cos²x-sin²x)-sinx=0

cos²x+sin²x-cos²x+sin²x-sinx=0

2sin²x-sinx=0

sinx*(2sinx-1)=0

1) sinx=0        

x=πn,   n∈Z  

2) 2sinx-1=0

2sinx=1

sinx=1/2

x(1)= π/6 +2πn,   n∈Z

x(x)= 5π/6 +2πn,   n∈Z

Ответ: πn, n∈Z;   π/6 +2πn, n∈Z;   5π/6 +2πn, n∈Z

Answer 2

cos²x cos 2x = cos²x (2cos²x - 1)
(1 - sin²x)(2cos²x - 1) = sin x
2cos²x - sin²x - 2cos²x sin²x + sin x = 0
sin²x + sin x - 1 = 0
x = (-b +/- sqrt(b² - 4ac)) / (2a)
x = (-1 +/- sqrt(1² - 4(1)(-1))) / (2(1))

x = (-1 +/- sqrt(1 + 4)) / 2

x = (-1 +/- sqrt(5)) / 2

x = (-1 + sqrt(5)) / 2 и x = (-1 - sqrt(5)) / 2