Question

а) Выбери формулы для корней уравнения 14sin2x+sin2x−10cos2x=1:
Варианты ответов:
1) arctg11/13,+πn,n∈Z
2) −arctg13/11,+πn,n∈Z
3) −arctg13/11,+2πn,n∈Z
4) −π/4+πk,k∈Z
5) π/4+πk,k∈Z
6) −arctg11/13,+2πn,n∈Z
7) другой ответ
8) −arctg11/13,+πn,n∈Z
9) −3π/4+πk,k∈Z
б) Посчитай количество корней, принадлежащих отрезку [3π;13π2].

Answer 1

$14Sin^{2} x+Sin2x-10Cos^{2} x=1\\\\14Sin^{2} x+Sin2x-10Cos^{2} x=Sin^{2}x+Cos^{2}x\\\\13Sin^{2}x+2SinxCosx-11Cos^{2}x=0 \ | \ :Cos^{2} x\neq 0\\\\\\\dfrac{13Sin^{2}x}{Cos^{2} x} +\dfrac{2SinxCosx}{Cos^{2} x}-\dfrac{11Cos^{2}x}{Cos^{2}x } =0\\\\\\13tg^{2}x+2tgx-11=0\\\\tgx=a\\\\13a^{2}+2a-11=0\\\\D=2^{2}-4\cdot13 \cdot(-11)=4+572=576=24^{2}\\\\a_{1} =\dfrac{-2-24}{26} =-1\\\\a_{2}=\dfrac{-2+24}{26}=\dfrac{22}{26}=\dfrac{11}{13} \\\\1) \ tgx=-1$

$\boxed{x=-\dfrac{\pi }{4} +\pi n \ ; \ n\in Z} \\\\\\2) \ tgx=\dfrac{11}{13} \\\\\boxed{x=arctg\dfrac{11}{13} +\pi n \ ; \ n\in Z}$

$1) \ 3\pi\leq -\dfrac{\pi }{4}+\pi n\leq \dfrac{13\pi }{2} \\\\12\pi \leq -\pi +4\pi n\leq 26\pi \\\\12\leq -1+4n\leq 26\\\\13\leq 4n\leq 27\\\\3,25\leq n\leq 6,75\\\\\boxed{3 \ kornia}\\\\\\2) \ 3\pi \leq arctg\dfrac{11}{13}+\pi n\leq \dfrac{13\pi }{2} \\\\3-\approx0,22\leq n \leq 6,5-\approx0,22 \\\\\approx2,78\leq n \leq \approx6,28 \\\\\boxed{4 \ kornia}\\\\Otvet:\boxed{7 \ kornei}$