Задача 8470 ...

Julia_Trusova

10.04.2016

а) Решите уравнение 1/(sin^2x)-3/(cos(3π/2-x))+2=0
б) Найдите все корни этого уравнения, принадлежащие отрезку [2π;7π/2]

а)1/(sin^2x)+3/(sinx)+2=0
ОДЗ: sinx≠0
х≠πn, nЄZ

(1+3sinx+2sin^2x)/(sin^2x)=0
1+3sinx+2sin^2x=0
Замена: sinx=t, -1≤t≤1
2t^2+3t+1=0
D=9-8=1
t1=(-3-1)/4=-1
t2=(-3+1)/4=-1/2

1)sinx=-1
x=-π/2+2πk, kЄZ

2)sinx=-1/2
x1=7π/6+2πl, lЄZ
x2=-π/6+2πm, mЄZ

Ответ: -π/2+2πk, kЄZ, 7π/6+2πl, lЄZ, -π/6+2πm, mЄZ

б)1) x=-π/2+2πk, kЄZ
2π≤-π/2+2πk≤7π/2
5π/2≤2πk≤4π
5/4≤k≤2
=>k=2, x=-π/2+2π*2==-π/2+4π=7π/2

2) x=7π/6+2πl, lЄZ
2π≤7π/6+2πl ≤7π/2
5π/6≤2πl ≤7π/3
5/12≤ l ≤7/6
=>l=1, x=7π/6+2π=19π/6

3) x2=-π/6+2πm, mЄZ
2π≤-π/6+2πm ≤7π/2
13π/6≤2πm ≤11π/3
13/12≤m ≤11/6
=>m=∅
Ответ: 7π/2, 19π/6

Ответ: а)-π/2+2πk, kЄZ, 7π/6+2πl, lЄZ, -π/6+2πm, mЄZ, б)7π/2, 19π/6