можно и так 7. (cos4x+1)sin2x =0 ; 2cos²2x*sin2x =0 ; a) cos2x =0 2x =π/2+π*n , n ∈ Z ; x =π/4+(π/2)*n , n ∈ Z. b) sin2x =0 ; 2x = π*n , n ∈ Z; x = (π/2)*n , n ∈ Z.
ответ : π/4+(π/2)*n , (π/2)*n , n ∈ Z. ------- 8. (cos(x+π/4) +1) *(tqx+1) =0 ; 2cos²(x/2+π/8) * (tqx+1) = 0 ; ⇔ (равносильно совокупности уравнений)[ cos(x/2+π/8)=0 ;tqx+1 =0. a) cos(x/2+π/8) =0 ; x/2+π/8 =π/2 +2π*n , n ∈ Z x =3π/4 +4π*n , n ∈ Z. b) tqx+1 = 0 ; tqx = -1 ; x = -π/4 +π*n , n ∈ Z.
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Verified answer
7) Cos4x + 1 = 0 или Sin2x = 0Cos4x = -1 2x = nπ , n ∈Z
4x = π + 2πk , k ∈ Z x = nπ/2, n ∈ Z
x = π/4 + πk/2 , k ∈Z
8) Cos(x + π/4) +1 = 0 или tgx +1 = 0
Cos(x +π/4) = -1 tgx = -1
x+π/4 = π + 2πk , k∈Z x = -π/4 + πn , n ∈Z
x = -π/4 + π + 2πk , k∈Z
x = 3π/4 + 2πk , k∈Z
Verified answer
можно и так7.
(cos4x+1)sin2x =0 ;
2cos²2x*sin2x =0 ;
a)
cos2x =0
2x =π/2+π*n , n ∈ Z ;
x =π/4+(π/2)*n , n ∈ Z.
b)
sin2x =0 ;
2x = π*n , n ∈ Z;
x = (π/2)*n , n ∈ Z.
ответ : π/4+(π/2)*n , (π/2)*n , n ∈ Z.
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8.
(cos(x+π/4) +1) *(tqx+1) =0 ;
2cos²(x/2+π/8) * (tqx+1) = 0 ; ⇔ (равносильно совокупности уравнений)[ cos(x/2+π/8)=0 ;tqx+1 =0.
a)
cos(x/2+π/8) =0 ;
x/2+π/8 =π/2 +2π*n , n ∈ Z
x =3π/4 +4π*n , n ∈ Z.
b)
tqx+1 = 0 ;
tqx = -1 ;
x = -π/4 +π*n , n ∈ Z.
ответ : 3π/4 +4π*n ; - π/4 +π*n , n ∈ Z.
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P.S. 1 +cosα =1+cos2*(α/2) = 1+cos²(α/2) - sin²(α/2) = 2cos²(α/2) .
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Удачи !