Task/27268281
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1.
cos²3x - sin²3x =0; * * * cos²α - sin²α =cos2α * * *
cos6x=0 ;
6x =π/2 +πn , n ∈Z.
x =π/12 +(π/6)*n , n ∈Z .
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2.
sinx*cosx = √3 /4 ; * * * 2sinα *cosα =sin2α * * *
sin2x =√3 /2 ;
2x = (-1)ⁿ *(π/3) +πn , n ∈Z .
x = (-1)ⁿ *(π/6) +(π/2)*n , n ∈Z .
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3.
5sin²x +sinx - 6 = 0 , замена t = sinx , -1≤ t ≤ 1
5t² + t - 6 =0 ; D = 1² -4*5*(-6) = 1+120 =121 =11²
t₁ = (-1-11)/ (2*5 ) = -1,2 < -1 , посторонний корень
t₂ =( -1+11) /10 = 1 .
обратная замена :
sinx = 1;
x = π/2 +2πn , n ∈ Z.
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4.
1 - cos2x = 2sinx ; * * * cos2x =cos²x -sin²x =1-2sin²x
2sin²x = 2sinx ;
sinx(sinx -1) =0 ;
sinx = 0 ⇒ x₁ =πn , n∈Z
или
sinx -1 =0 ⇔sinx=1 ⇒ x₂=π/2 +2πn , n∈Z.
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5 .
5sin²x + 6cosx - 6 = 0 ;
5(1 -cos²x)+6cosx - 6 = 0 ;
5cos²x - 6cosx + 1 = 0 ; D₁ = 3² -5*1 =4 =2² ; * * * кв. урав. отн. cosx * * *
cosx₁ =(3 -2)/5 =1/5 ⇒ x₁ =(-1)ⁿ *arcsin(1/5) +πn , n ∈Z .
cosx₂ =(3 +2)/5 =1 . ⇒ x₂ = 2πn , n ∈Z .