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aruzina
@aruzina
July 2022
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Решите пожалуйста 1) 3-4соs^2x=0 2) 5 sin^2x-9 sinx+4=0 3) 3sinx-7cosx=0 4) 3sin^2x-8sinxcosx+5cos^2x=0
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oganesbagoyan
Verified answer
1)
3 - 4cos²x =0 ;
3 -4* (1+cos2x)/2 =0 ;
3 -2 -cos2x =0 ;
cos2x =1 ⇒2x=2πn ,n∈Z⇔
x=πn ,n∈Z.
-------
2)
5sin²x - 9sinx +4 =0 ;
5t² -9t +4 =0 ;
D =9² -4*5*4 =1.
t₁ =1 ⇒sinx =1⇒
x₁ =π/2 +2πn ,n∈Z.
t₂ =4/5 ⇒sinx =4/5⇒
x₂ =(-1)^n +πn ,n∈Z.
-------
3)
3sinx -7cosx =0 ;
sinx =(7/3)cosx ; * * * разделим на cosx ≠0 * * *
* если cosx =0 ⇒sinx =0 противоречие sin²x +cos²x =1 * * *
tqx =7/3 ;
x =arctq(7/3) + πn ,n∈Z.
-------
4)
3sin²x -8sinxcosx +5cos²x =0 ; || :cos²x
3tq²x -8tqx +5 =0; * * * замена t =
tqx * * *
3t² -8t +5 =0 ;
D =4² -3*5=1.
t₁ =(4-1)/3 =1⇒tqx₁ =1⇒
x₁ =π/4 +π*n , n∈Z.
t₂=(4+1)/3 =5/3 ⇒tqx₂ =5/3⇒
x₂ = arctq(5/3) +π*n , n∈Z.
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aruzina
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Answers & Comments
Verified answer
1)3 - 4cos²x =0 ;
3 -4* (1+cos2x)/2 =0 ;
3 -2 -cos2x =0 ;
cos2x =1 ⇒2x=2πn ,n∈Z⇔ x=πn ,n∈Z.
-------
2)
5sin²x - 9sinx +4 =0 ;
5t² -9t +4 =0 ;
D =9² -4*5*4 =1.
t₁ =1 ⇒sinx =1⇒ x₁ =π/2 +2πn ,n∈Z.
t₂ =4/5 ⇒sinx =4/5⇒ x₂ =(-1)^n +πn ,n∈Z.
-------
3)
3sinx -7cosx =0 ;
sinx =(7/3)cosx ; * * * разделим на cosx ≠0 * * *
* если cosx =0 ⇒sinx =0 противоречие sin²x +cos²x =1 * * *
tqx =7/3 ;
x =arctq(7/3) + πn ,n∈Z.
-------
4)
3sin²x -8sinxcosx +5cos²x =0 ; || :cos²x
3tq²x -8tqx +5 =0; * * * замена t = tqx * * *
3t² -8t +5 =0 ;
D =4² -3*5=1.
t₁ =(4-1)/3 =1⇒tqx₁ =1⇒ x₁ =π/4 +π*n , n∈Z.
t₂=(4+1)/3 =5/3 ⇒tqx₂ =5/3⇒ x₂ = arctq(5/3) +π*n , n∈Z.