Question

Решите!!!!!
8sinx+cosx=4

Answer 1

Пусть a = x/2
sin(x) = 2sin(a)cos(a);
cos(x) = cos^2(a) - sin^2(a)
1/cos^2(x) = 1 + tg(a)

8sin(x) + cos(x) = 4
8*2sin(a)cos(a) + cos^2(a) - sin^2(a) = 4

Разделим обе части уравнения на cos^2(a):
16*tg(a) + 1 - tg^2(a) = 4*(1 + tg^2(a))
4 tg^2(a) + tg^2(a) + 3 - 16 tg(a) = 0
5 tg^2(a) - 16 tg(a) + 3 =0

D = 16^2 - 4*5*3 = 196 = 14^2
tg(a) = (16 + 14) / 10; tg(a) = (16-14)/10
tg(a) = 3;                     tg(a) = 1/5;
a = arctg(3) + πn, n∈Z;          a = arctg(1/5) + πk, k∈Z

x/2 = arctg(3) + πn, n∈Z;       x/2 = arctg(1/5) + πk, k∈Z;
x = 2 arctg(3) + 2 πn, n∈Z;        x = arctg(1/5) + 2 πk, k∈Z;

Ответ: 2 arctg(3) + 2 πn, n∈Z; arctg(1/5) + 2 πk, k∈Z;