Question

помогите решить!!!

sin2x-22cos^2x+10=0

Answer 1

7sin2x-22cos^2x+10=0

14sinxcosx - 22cos^2x + 10(sin^2x + cos^2x) = 0

14sinxcosx -12cos^2x +10sin^2x = 0  |:cos^2x

14tgx - 12 + 10tg^2x = 0

5tg^2x +7tgx - 6 = 0

tgx = t

5t^2 +7t - 6 = 0

D = 49 + 4*5*6 = 169

t = (-7 +- 13)/10

t = -2   t = 3/5

tgx = -2                            tgx = 3/5

x = -arctg2 + Пk                x = arctg(3/5) + Пk

Answer 2

Ответ:

x1 = pi/4 + pi*n, 

x2 = -arctg(1,2) + Pi*m

m, n принадлежат Z