1)7sin2y=2siny 2)(3cosx+8sinx)^2=12+55sin^2x

1) 7*sin2y = 2siny 7*(2siny*cosy) - 2siny = 0 7*siny*cosy - siny = 0 siny*(7cosy - 1 ) = 0 siny = 0 ==> y = pik, k ∈z cosy = 1/7 ==> y = ± arсcos(1/7) + 2pik,  k ∈z 2) (3cosx + 8sinx)^2 = 12 + 55sin^2x 9cos^2x + 48sinxcosx + 64sin^2x - 12(sin^2x + cos^2x)  - 55sin^2x = 0 9cos^2x + 48sinxcosx + 64sin^2x - 12sin^2x - 12cos^2x  - 55sin^2x = 0 - 3 sin^2x + 48sinxcosx  - 3 cos^2x = 0   (sin^2x + cos^2x) - 16sinxcosx  = 0   1 – 8sin2x= 0  sin2x = 1/8  2x = arcsin(1/8) + 2pik x = 1/2 * arcsin(1/8) + pik ,  k ∈z x = pi/2 - 1/2* arcsin(1/8) + pik ,  k ∈z

Y`=-2x sin(x-1) y``=-2cos(x-1) n> 2 y(n)=-2sin(x-1+pi*(n-1)/2)

x^2-6x+5=0

d=9-5=4

x1=3+2=5

x2=3-2=1

ответ: 1; 5