Question

cos 3x + cos x =0 промежуток [-П2;П2]
и
2 sin в крадрате x - sin 2x=cos 2x

Заранее спасибо)

Answer 1

1)  cos3x + cosx = 0
2cos(3x + x)/2*cos(3x - x)/2 = 0
cos2x * cosx = 0
a)  cos2x = 0
2x = π/2 + πn, n∈Z
x₁ = π/4 + πn/2, n ∈Z
n = - 1
x = π/4 - π/2 = - π/4 ∈ [- π/2;π/2]
n = - 2
x = π/4 - π = - 3π/4 ∉ [- π/2;π/2]
n = 0
x = π/4 ∈ [- π/2;π/2]
n = 1
x = π/4 + π/2 = 3π/4 ∉ [- π/2;π/2]
n = 2 
π/4 + π = 5π/4 ∉ [- π/2;π/2]
Ответ: - π/4;  π/4
b)  cosx = 0
x = π/2 + πk, k∈Z
k = - 1
x = π/2 - π = - π/2 ∈ [- π/2;π/2]
k = 0
x = π/2  ∈ [- π/2;π/2]
k = 1
x = π/2 + π = 3π/2 ∉  [- π/2;π/2]
Ответ: - π/2; π/2 

2) 2sin² x - sin2x = cos2x
2sin²x - 2sinxcosx - (2cos²x - 1) = 0
2sin²x - 2sinxcosx - 2cos²x + sin²x + cos²x = 0
3sin²x - 2sinxcosx – cos²x = 0    / делим на cos²x ≠ 0
3tg²x - 2tgx - 1 = 0
D = 4 + 4*3*1 = 16
1) tgx = (2 - 4)/6
tgx = - 1/3
x₁ = - arctg(1/3) + πn, n∈Z
tgx = ( 2 + 4)/6
tgx = 1 
x₂ = π/4 + πk, k∈Z