Question

1)7sin2y=2siny
2)(3cosX+8sinX)^2=12+55sin^2X

Answer 1

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

Answer 2

представили 12*1
основное тригонометрическое тождество

Answer 3

sin^2x +

Answer 4

cos^2x