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DainmaarDovah
@DainmaarDovah
August 2022
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Помогите, пожалуйста, с примерами.
sin3x*cosx+sinx*cos3x=0
sinx+[tex] \sqrt{3} [/tex]cosx=0
sin2xcosx-2sinx=0
[tex] {sin^{2}x } [/tex]+3sinx*cosx-4[tex] {cos^{2}x } [/tex]=0
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oganesbagoyan
Verified answer
Task/25107582
----------------------
* * * sin(α+β) =sinα*cosβ+cos
α*sinβ * * *
sin3x*cosx+sinx*cos3x=0⇔sin4x=0⇒ 4x =π*n ,n∈Z .
ответ : x =(π/4)*n , n∈Z.
-----------sinx+ (√3)* cosx=0 ⇔sinx= - (√3)*cosx ⇔ tgx = -√3 * * *т.к. cosx ≠ 0 * * *
ответ : x = -π/3 + π*n ,n∈Z .
-----------
* * *sin2x =2sinx*cosx * * *
sin2xcosx-2sinx=0⇔2sinx*cosx*cosx -2sinx=0 ⇔2sinx(cos²x -1) =0 ⇔-
2sin³x =0 ⇔sinx=0 ;
ответ : x = π*n ,n∈Z.
-----------
sin²x +3sinx*cosx- 4cos²x= 0 ⇔tg²x +3tgx -4 =0 ⇒[ tgx = - 4 , tgx =1 .
ответ: x = -arctg4 +π*n , n∈Z , x =π/4 +π*k ,k∈Z .
============
Удачи !
2 votes
Thanks 1
DainmaarDovah
огромное спасибо!
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Verified answer
Task/25107582----------------------
* * * sin(α+β) =sinα*cosβ+cosα*sinβ * * *
sin3x*cosx+sinx*cos3x=0⇔sin4x=0⇒ 4x =π*n ,n∈Z .
ответ : x =(π/4)*n , n∈Z.
-----------sinx+ (√3)* cosx=0 ⇔sinx= - (√3)*cosx ⇔ tgx = -√3 * * *т.к. cosx ≠ 0 * * *
ответ : x = -π/3 + π*n ,n∈Z .
-----------
* * *sin2x =2sinx*cosx * * *
sin2xcosx-2sinx=0⇔2sinx*cosx*cosx -2sinx=0 ⇔2sinx(cos²x -1) =0 ⇔-
2sin³x =0 ⇔sinx=0 ;
ответ : x = π*n ,n∈Z. -----------
sin²x +3sinx*cosx- 4cos²x= 0 ⇔tg²x +3tgx -4 =0 ⇒[ tgx = - 4 , tgx =1 .
ответ: x = -arctg4 +π*n , n∈Z , x =π/4 +π*k ,k∈Z .
============
Удачи !