Решить поcледовательно, а не просто ответ.

19sin2x + 6cos^2x - 12 = 0;

9cosx + sinx - 1 = 0;

19sin2x + 6cos^2x - 12 = 0

19*2sinxcosx+6cos^2x - 12 (cos^2x+sin^2x) = 0

-12sin^2x+38sinxcosx-6cos^2x=0 / : (-2)

6sin^2x-19sinxcosx+3cos^2x=0 / : cos^2x≠0

6tg^2x-19tgx+3=0

tgx=t

6t^2-19t+3=0

D=289

t=3

t=1/6

x=arctg (3) + pik, k∈Z

x=arctg (1/6) + pik, k∈Z

9cosx + sinx - 1 = 0

подстановка tg (x/2) = t

9 * (1-t^2) / (1+t^2) + 2t / (1+t^2) - 1=0

9 * (1-t^2) + 2t - (1+t^2) = 0

9-9t^2+2t-1-t^2=0

-10t^2+2t+8=0 / : (-2)

5t^2-t-4=0

D=1+16*5=81

t = (1+9) / 10=1

t = (1-9) / 10=-8/10=-4/5

tg (x/2) = 1

x/2=pi/4+pik / * 2

x=pi/2+2pik

tg (x/2) = - 4/5

x/2=-arctg (4/5) + pik

x=-2arctg (4/5) + 2pik