249(1,3) 250(1,3) Спасибо !!!!!. Created by stiklin. algebra-ru

249(1,3) 250(1,3) Спасибо !!!!!...

stiklin @stiklin

August 2022 1 34 Report

249(1,3)
250(1,3)
Спасибо !!!!!

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oganesbagoyan

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Task/26534690
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249. 1)
250(1,3)2cos(x/2) = 1+cos x ;
2cos(x/2) =2cos² (x/2) ;
2cos(x/2)*(cos(x/2) -1) = 0 ;
а)
cos(x/2) =0 ;
x/2=  π/2 +πk , k ∈ ℤ ;
x = π +2πk , k ∈ ℤ .
б)
cos(x/2) -1 = 0 ;
cos(x/2) = 1 ;
x/2=  2πn , n ∈ ℤ ;
x = 4πn , n ∈ ℤ .
x = 2πk , k ∈ ℤ .

ответ : 2πk ; π +2πk , k ∈ ℤ .
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249. 3)
sin3x / sinx = 0 ;
sinx (3 - 4sin²x) / sinx = 0 ; * * *ОДЗ: sinx ≠ 0* * *
3 - 4sin²x = 0 ;
3 - 4*(1-cos2x)/2 =0 ;
3 - 2 +2cos2x =0 ;
cos2x = - 1/2 ;
2x = ±(π-π/3) +2πn , n∈Z  ;
2x = ±2π/3 +2πn , n∈Z  ;
x = ±π/3 + πn , n∈Z .

ответ :  ±π/3 + πn , n∈Z  .
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250. 1)
2sin²x +sinx =1 ;
2sin²x +sinx -1=0 ; квадратное уравнение относительно t =sinx
sinx₁ =(-1 -3)/4= -1 ⇒ x₁ = -π/2 +πn , n∈ℤ;
sinx₂=(-1 +3)/4=1/2 ⇒ x₂ = (-1)^n *π/6 +πn , n∈ℤ.

ответ :  -π/2 +πn ;   (-1)^n *π/6 +πn , n∈Z.  n∈ℤ  .
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250. 3)
4sin⁴x +cos4x =1 +12cos⁴x ;
* * * имеем : sin²α =(1-cos2α)/2 ; cos²α =(1+cos2α)/2 * * *
4*( (1-cos2x)/2 )²+cos4x = 1 +12*( (1+cos2x)/2 )² ;
(1- cos2x)² + cos4x=1 +3(1+cos2x)² ;
1 - 2cos2x +cos²2x +cos4x =1 +3(1 +2cos2x +cos²2x);
1 +cos4x +cos²2x - 2cos2x =3cos²2x +6co2x +4 ;
* * *1 +cos4x= 2cos²2x * * *
2cos²2x +cos²2x - 2cos2x =3cos²2x +6co2x +4 ;
8cos2x = - 4 ;
cos2x = -1/2 ; уже решен в  249. 3)
x = ±π/3 + πn , n∈Z .

ответ :  ± π/3 + πn , n∈Z  .
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Удачи !

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