Решите уравнение
а)2cosx -2cos²x +sin²x =0 ;
б) Найдите все корни этого уравнения принадлежащие отрезку [ 3π ; 9π/2 ]
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а)
2cosx -2cos²x +1 -cos²x =0 ;
3cos²x -2cosx -1 =0 ;
* * *3cos²x -3cosx +cosx -1 =3cosx(cosx - 1) +(cosx -1) =(cosx - 1)(3cosx +1) * *
[ cosx = 1 ; cosx = -1/3
или стандартно, замена: cosx =t
3t² -2t -1 =0 ; D/4 =(2/2)² -3*(-1) =4 =2² * * * D =16 * * *
t₁= (1+2) /3 =1 ;
t₂ =(1-2) /3 = - 1/3.
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а₁)
cosx =1 ;
x =2πn , n ∈ Z.
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а₂)
cosx = -1/3 ;
x = ± ( π -arccos(1/3) ) +2πk , k ∈ Z.
ответ: 2πn , n ∈ Z и ± ( π -arccos(1/3) ) +2πk , k ∈ Z
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
б) x ∈[ 3π ; 9π/2 ]
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б₁) x =2πn , n ∈ Z.
3π ≤ 2πn ≤ 9π/2⇔ 3/2 ≤ n ≤ 9/4 ⇒ n =2 ,
т.е. x =4π .
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б₂) x = ± ( π -arccos(1/3) ) +2πk , k ∈ Z. разделяем
б₂₁)
3π ≤ - π +arccos(1/3) +2πk ≤ 9π/2 ;
4π - arccos(1/3) ≤ 2πk ≤ 11π/2 -arccos(1/3)
2 - arccos(1/3) / 2π ≤ k ≤ 11/4 -arccos(1/3) / 2π ⇒ k =2 ,
т.е. x = 3π +arccos(1/3)
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б₂₂)
3π ≤ π -arccos(1/3) +2πk ≤ 9π/2 ;
2π +arccos(1/3) ≤ 2πk ≤ 7π/2 +arccos(1/3) ;
1 +arccos(1/3) / 2π ≤ k ≤ 7/4 +arccos(1/3) / 2π ⇒ k∈∅
ответ: 3π +arccos(1/3) , 4π .
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Удачи !