q=1,2/0,4=3 значит b3=1,2*3=3,6 b4=3,6*3=10,8 b5=10,8*3=34,4
(7sin α - 2cos α)/(4sin α - cos α) = 2
Делим числитель и знаменатель дроби на cos α ≠ 0
(7tg α - 2)/(4tg α - 1) = 2 ОДЗ: tg α ≠ 1/4
Умножаем левую и правую части уравнения на (4tg α - 1)
(7tg α - 2) = 2(4tg α - 1)
7tg α - 2 = 8tg α - 2
8tg α - 7tg α = -2 + 2
tg α = 0
Ответ: tg α = 0
7) =-1/2* ln(3 - 2x)|₋₁¹ = -1/2*ln(3 -2*1) -(-1/2*ln(3 -2*(-1)) = -1/2*ln1 + 1/2ln5 =0,5 ln5
8) = - 6Cosx/6+1/5Sin5x)| от 0 до 2π =
= -6*Сos2π/3 + 1/5Sin10π - (-6Cos0 + 1/5Sin0)= 3 +1/*0 + 6 -1/5*0 = 9
9) = 1/3*Cos(π/3 -3x)| в пределах от 0 до 2π =
= 1/3*Сos(π/3 -3*2π) - 1/3*Cos(π/3 -0)= 1/3Cosπ/3 -1/3Cosπ/3 = 0
Ответ:
x∈(-3;-2]U[1;2)
Объяснение:
![\displaystyle\\\sqrt{x^2+x-2} <2\\\\\\\left \{ {{x^2+x-2\geq 0 } \atop {x^2+x-2<2^2}}\right.\\\\\\\left \{ {{x^2+x-2\geq 0 } \atop {x^2+x-6<0}}\right.\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%5C%5C%5Csqrt%7Bx%5E2%2Bx-2%7D%20%3C2%5C%5C%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7Bx%5E2%2Bx-2%5Cgeq%200%20%7D%20%5Catop%20%7Bx%5E2%2Bx-2%3C2%5E2%7D%7D%5Cright.%5C%5C%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7Bx%5E2%2Bx-2%5Cgeq%200%20%7D%20%5Catop%20%7Bx%5E2%2Bx-6%3C0%7D%7D%5Cright.%5C%5C%5C%5C)
по теореме Виета
![\displaystyle\\\left \{ {{(x-1)(x+2)\geq0 } \atop {(x-2)(x+3)<0}} \right.\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%5C%5C%5Cleft%20%5C%7B%20%7B%7B%28x-1%29%28x%2B2%29%5Cgeq0%20%7D%20%5Catop%20%7B%28x-2%29%28x%2B3%29%3C0%7D%7D%20%5Cright.%5C%5C%5C%5C%5C%5C)
решаем методом интервалов
++++++++[-2]-----------[1]++++++++++
++++(-3)-------------------------(2)+++++
![\displaystyle\\\left \{ {{x\in(-\infty;-2]U[1;+\infty)} \atop {x\in(-3;2)}} \right.\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%5C%5C%5Cleft%20%5C%7B%20%7B%7Bx%5Cin%28-%5Cinfty%3B-2%5DU%5B1%3B%2B%5Cinfty%29%7D%20%5Catop%20%7Bx%5Cin%28-3%3B2%29%7D%7D%20%5Cright.%5C%5C%5C%5C%5C%5C)
ответ: x∈(-3;-2]U[1;2)