Task/26598992
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256. 4)
3sinx + 4cosx +5sin3x =0 ;
можно использовать метод вспомогательного угла )¹
* * * 3sinx + 4cosx = 5sin(x + φ) .* * *
5sin(x + φ) +5sin3x =0 , где φ =arctg( 4/3)
5(sin3x +sin(x+φ) ) =0 ;
2sin(2x +φ/2)*cos(x - φ/2) =0 ;
а)
sin(2x +φ/2) = 0 ;
2x +φ/2 =πn , n ∈ Z ;
x = - φ/4 +(π/2)* n , n ∈ Z ;
x = - (1/4)arctg(4/3) +(π/2)* n , n ∈ Z ;
б)
cos(x - φ/2) =0 ;
x - φ/2 = π/2 +πn , n ∈ Z ;
x = φ/2 +π/2 +πn , n ∈ Z ;
x =(1/2)*arctg(4/3) + π/2 +πn , n ∈ Z .
ответ : - (1/4)arctg(4/3) +(π/2)* n , (1/2)*arctg(4/3) + π/2 +πn , n ∈ Z .
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257. 1)
(sin7x +cos7x)² = 2sin²11x ;
sin²7x +2sin7x*cos7x +cos²7x =2*(1-cos22x) / 2 ;
1 +sin2*(7x) = 1 - cos22x ;
cos22x +sin14x =0 ;
cos22x +cos(π/2 - 14x) =0 ;
2cos(4x +π/4)*cos(18x -π/4) =0 ;
а)
cos(4x +π/4) =0 ;
4x +π/4 =π/2 +πn , n ∈ Z ;
x = π / 16 + (π/4)* n , n ∈ Z .
б)
cos(18x -π/4) =0:
18x -π/4 = π/2 +πn , n ∈ Z ;
x = π/24 +(π/18)*n , n ∈Z .
ответ : π / 16 + (π/4)* n , π/24 +(π/18)*n , n ∈Z .
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)¹
метод вспомогательного угла :
asinx +bcosx =√(a² +b²)* (sinx *a /√(a² +b²) +cosx*b /√(a² +b²) )=
√(a² +b²)* (sinx *cosφ +cosx*sinφ) =√(a² +b²) sin(x +φ) .
b /√(a² +b²) =sinφ ;
a /√(a² +b²) =cosφ ;
tgφ = b/a ⇒ φ =arctg (b/a)
* * * 3sinx + 4cosx =√(3² +4²)sin(x +φ) =5sin(x+ φ) ; arctg(4/3) * * *