Question

cos^2 2x+ cos^2 x + cos^2 3x + cos^2 4x = 2 Пожалуйста, помогите решить.....

Answer 1

cos²x + cos²2x + cos²3x + cos²4x = 2

cos2x = 2cos²x - 1  ⇒  cos²x = ( 1 + cos2x )/2

(1 + cos2x)/2  +  (1 + cos4x)/2  +  (1 + cos6x)/2  + (1 + cos8x)/2 = 2

1 + cos2x + 1 + cos4x + 1 + cos6x + 1 + cos8x = 4

(cos2x + cos8x) + (cos4x + cos6x) = 0

cosα + cosβ = 2•cos( (1/2)•(α + β) )•cos( (1/2)•(α - β) )

2•cos5x•cos3x + 2•cos5x•cosx = 0

2•cos5x•(cos3x + cosx) = 0

2•cos5x•2•cos2x•cosx = 0

4•cos5x•cos2x•cosx = 0

1) cos5x = 0  ⇒  5x = (π/2) + πn  ⇒  x = (π/10) + (πn/5) , n ∈ Z

2) cos2x = 0  ⇒  2x = (π/2) + πk  ⇒  x = (π/4) + (πk/2) , k ∈ Z

3) cosx = 0  ⇒  x = (π/2) + πm , m ∈ Z

ОТВЕТ: (π/10) + (πn/5) , (π/4) + (πk/2) , (π/2) + πm , n,k,m ∈ Z