Задача 12383 ...

SOVA

19.12.2016

А) Решите уравнение cos2x+sqrt(2)*sin((π/2)–x)+1=0
Б) Найдите всё корни этого уравнения , принадлежащие отрезку [–4π;–5π/2]

По формулам приведения
sin((п/2)–x)=cosx
По формуле двойного угла
сos2x=2cos2x–1
Уравнение принимает вид:
2сos2x+√2cosx=0
cosx·(2cosx+√2)=0
cosx=0 или cosx=–√2/2
x=(π/2)+πk, k∈Z или х=±(3π/4)+2πn, n∈Z

Б)–4π≤(π/2)+πk≤–5π/2⇒
–4 ≤(1/2)+k ≤–5/2
–4,5 ≤ k ≤–3
k=–4; k=–3
x=(π/2)–4π=–7π/2∈[–4π;–5π/2];
x=(π/2)–3π=–5π/2∈[–4π;–5π/2];

–4π≤(3π/4)+2πn≤–5π/2⇒
–4 ≤(3/4)+2n ≤–5/2;
–4целых3/4 ≤2n ≤(–5/2)–(3/4);
–19/4 ≤2n ≤–13/4;
–19 ≤8n ≤–13;
n=–2
x=(3π/4)–4π=–13π/4∈[–4π;–5π/2]


–4π≤(–3π/4)+2πn≤–5π/2⇒
–4 ≤(–3/4)+2n ≤–5/2;
–4+(3/4) ≤2n ≤(–5/2)+(3/4);
–13/4 ≤2n ≤–7/4;
–13 ≤8n ≤–7;
n=–1
x=(–3π/4)–2π=(–11π/4)∈[–4π;–5π/2]

О т в е т. A)(π/2)+πk, ±(3π/4)+2πn, k, n∈Z
Б)–7π/2;–13π/4;–11π/4;–5π/2 – корни, принадлежащие[–4π;–5π/2];

Ответ: A)(π/2)+πk, ±(3π/4)+2πn, k, n∈Z Б)–7π/2;–13π/4;–11π/4;–5π/2 – корни, принадлежащие[–4π;–5π/2];