Решите!

8sinx+cosx=4

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;