Sin 3x - Sin 7x = √<span>3 Sin 2x
-2Cos5x Sin2x = </span>√3Sin2x
-2Cos5x Sin2x - √3Sin2x = 0
2Cos5x Sin2x + √3Sin2x = 0
Sin2x(2Cos5x + √3) = 0
Sin2x = 0 2Cos5x +√3 = 0
2x = πn , n∈Z Cos5x = -√3/2
x = nπ/2, n ∈ Z 5x = +-arcCos(√3/2) + 2πk , k ∈Z
5x = +-π/6 + 2πk , k ∈Z
x = +-π/30 + 2πk/2, k ∈Z
Так как
, то по теореме Виета
![x_1+x_2=p~~\Rightarrow~~ x_1=p-x_2](https://tex.z-dn.net/?f=x_1%2Bx_2%3Dp~~%5CRightarrow~~+x_1%3Dp-x_2)
![x_1x_2=-10~~\Rightarrow~~~ (p-x_2)x_2=-10](https://tex.z-dn.net/?f=x_1x_2%3D-10~~%5CRightarrow~~~+%28p-x_2%29x_2%3D-10)
И решим уравнение
в целых числах.
Делители числа 10: 1, 2, 5, 10.
![\displaystyle \left \{ {{p-x_2=1} \atop {x_2=-10}} \right.~~\Leftrightarrow~~~\left \{ {{p=-9} \atop {x_2=-10}} \right.\\ \\ \left \{ {{p-x_2=-10} \atop {x_2=1}} \right.~~\Rightarrow~~~\left \{ {{p=-9} \atop {x_2=1}} \right.\\ \\ \left \{ {{p-x_2=-1} \atop {x_2=10}} \right.~~\Rightarrow~~\left \{ {{p=9} \atop {x_2=10}} \right.\\ \left \{ {{p-x_2=10} \atop {x_2=-1}} \right.~~\Rightarrow~~\left \{ {{p=9} \atop {x_2=-1}} \right.](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D1%7D+%5Catop+%7Bx_2%3D-10%7D%7D+%5Cright.~~%5CLeftrightarrow~~~%5Cleft+%5C%7B+%7B%7Bp%3D-9%7D+%5Catop+%7Bx_2%3D-10%7D%7D+%5Cright.%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D-10%7D+%5Catop+%7Bx_2%3D1%7D%7D+%5Cright.~~%5CRightarrow~~~%5Cleft+%5C%7B+%7B%7Bp%3D-9%7D+%5Catop+%7Bx_2%3D1%7D%7D+%5Cright.%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D-1%7D+%5Catop+%7Bx_2%3D10%7D%7D+%5Cright.~~%5CRightarrow~~%5Cleft+%5C%7B+%7B%7Bp%3D9%7D+%5Catop+%7Bx_2%3D10%7D%7D+%5Cright.%5C%5C+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D10%7D+%5Catop+%7Bx_2%3D-1%7D%7D+%5Cright.~~%5CRightarrow~~%5Cleft+%5C%7B+%7B%7Bp%3D9%7D+%5Catop+%7Bx_2%3D-1%7D%7D+%5Cright.)
![\displaystyle \left \{ {{p-x_2=2} \atop {x_2=-5}} \right.~~\Rightarrow~~\left \{ {{p=-3} \atop {x_2=-5}} \right.\\ \\\left \{ {{p-x_2=-5} \atop {x_2=2}} \right.~~\Rightarrow~~\left \{ {{p=-3} \atop {x_2=2}} \right.\\ \\ \left \{ {{p-x_2=-2} \atop {x_2=5}} \right.~~\Rightarrow~~\left \{ {{p=3} \atop {x_2=5}} \right.\\ \\ \left \{ {{p-x_2=5} \atop {x_2=-2}} \right.~~\Rightarrow~~\left \{ {{p=3} \atop {x_2=-2}} \right.](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D2%7D+%5Catop+%7Bx_2%3D-5%7D%7D+%5Cright.~~%5CRightarrow~~%5Cleft+%5C%7B+%7B%7Bp%3D-3%7D+%5Catop+%7Bx_2%3D-5%7D%7D+%5Cright.%5C%5C+%5C%5C%5Cleft+%5C%7B+%7B%7Bp-x_2%3D-5%7D+%5Catop+%7Bx_2%3D2%7D%7D+%5Cright.~~%5CRightarrow~~%5Cleft+%5C%7B+%7B%7Bp%3D-3%7D+%5Catop+%7Bx_2%3D2%7D%7D+%5Cright.%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D-2%7D+%5Catop+%7Bx_2%3D5%7D%7D+%5Cright.~~%5CRightarrow~~%5Cleft+%5C%7B+%7B%7Bp%3D3%7D+%5Catop+%7Bx_2%3D5%7D%7D+%5Cright.%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7Bp-x_2%3D5%7D+%5Catop+%7Bx_2%3D-2%7D%7D+%5Cright.~~%5CRightarrow~~%5Cleft+%5C%7B+%7B%7Bp%3D3%7D+%5Catop+%7Bx_2%3D-2%7D%7D+%5Cright.)
Ответ: ± 3; ± 9.
<span>12,25 - 3х^2 = 6x^2
</span>12.25=9x²
x²=12.25/9
x=3.5/3=7/6
x=-7/6
((1/5)^2)^(-3/2) * ((1/2)^3)^(-4/3) - ((1/11)^2)^(-1/2) = (1/5)^(-3) * (1/2)^(-4)- (1/11)^(-1)=
=(5^3) * (2^4) - 11 = 125*16 - 11 = 1989