5.
2sin²(4x+π/4) + sin(4x+π/4) - 1 =0 ;
квадратное уравнение относительно sin(4x+π/4) ; D =1²-4*2*(-1)=3² ;
a)
sin(4x₁+π/4) = (-1 -3)/2*2 = - 1 ⇒ 4x₁ +π/4 = -π/2 +2πn , n ∈Z ;
x₁ = -3π/16 +πn/2 , n ∈Z;
b)
sin(4x+π/4) = (-1 +3)/2*2 = 1/2 ⇒
* * * 4x+π/4 = (-1)^n*(π/6)+πn , n ∈Z * * *
b₁) 4x₂ +π/4 = π/6+2πn , n ∈Z ⇔ x₂ = -π/48 +πn/2 , n ∈Z ;
b₂) 4x₃+π/4 = π -π/6+2πn , n ∈Z ⇔ x₃= 7π/12 +πn/2 , n ∈Z .
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6.
4sinxcosx = √2 ; * * * 2sinxcosx =sin2x* * *
sin2x =√2 /2 ;
2x = (-1)^n *π/4 +πn , n ∈Z ⇔x = (-1)^n *π/8 +πn/2 , n ∈Z.
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7.
sin²x -cos²x = 1 ;
cos²x - sin²x = - 1 ; * * * cos²x - sin²x = cos2x * * *
cos2x = -1 ;
2x =π+2πn , n ∈Z⇔x =π/2+πn, n ∈Z.
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8.
4sin²x = 3 + 2sinxcosx; * * * 3 = 3(sin²x +cos²x) =3sin²x +3cos²x * * *
sin²x -2sinxcosx -3cos²x =0 ;|| : cos²x ≠0
tg²x -2tgx -3 =0 ; D/4 =1² -(-3) =4 =2²
tgx₁=1 - 2 = -1 ⇒ x₁= -π/4 +πn, n ∈Z;
tgx₂=1 +2 =3 ⇒ x₂= arctg(3) +πn, n ∈Z.
* * * * * * * * * * * P.S. * * * * * * * * * * * *
sin²x -2cosx*sinx -3cos²x =0
sinx₁ =cosx₁ - 2cosx₁ = - cosx₁ ⇒tgx₁ = -1 ;
sinx₂ = cosx₂ +2cosx₂ = 3cosx₂ ⇒ tgx₂ = 3.
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2sin²(4x+π/4)+sin(4x+π/4) -1 =2sin²(4x+π/4)+2sin(4x+π/4)-sin(4x+π/4) - 1=
2sin(4x+π/4) *(sin(4x+π/4) + 1)- (sin(4x+π/4) + 1)=
2( sin(4x+π/4) + 1 )*( sin(4x+π/4) - 1/2 ).