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Помеха7
@Помеха7
July 2022
1
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Решите, пожалуйста, уравнения:
а) [tex](cos 2x - 1) * \sqrt{9 - x^{2} } = 0[/tex]
б) [tex]tgx * \sqrt{16 - x^{2} } = 0[/tex]
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oganesbagoyan
Verified answer
а)
(cos2x -1)*sqrt(9-x²)=0
а₁)
9-x²=0
x=(+∨-)3
а₂)
cos2x -1=0 и 9-x² ≥0
cos2x =1 и x∈ [-3;3]
2x=2π*k и x∈ [-3;3]
x=π*k и x∈ [-3;3]
x=0
(только при k =0)
ответ:
{- 3; 0 ; 3}
б)
tqx*sqrt(16-x²)=0
б₁)
16-x² =0 и tqx
x=(+∨-)4
б₂)
tqx =0 и 16-x² ≥0
x=π*k и x∈ [-4;4]
x= -π ; 0 ; π при k = -1 ,0,1
ответ:
{ -4 ; -π ; 0 ; π ; 4}
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Answers & Comments
Verified answer
а) (cos2x -1)*sqrt(9-x²)=0а₁) 9-x²=0
x=(+∨-)3
а₂) cos2x -1=0 и 9-x² ≥0
cos2x =1 и x∈ [-3;3]
2x=2π*k и x∈ [-3;3]
x=π*k и x∈ [-3;3]
x=0 (только при k =0)
ответ: {- 3; 0 ; 3}
б) tqx*sqrt(16-x²)=0
б₁)16-x² =0 и tqx
x=(+∨-)4
б₂) tqx =0 и 16-x² ≥0
x=π*k и x∈ [-4;4]
x= -π ; 0 ; π при k = -1 ,0,1
ответ: { -4 ; -π ; 0 ; π ; 4}