::::::::::::решение:::::::::::
3sin²+3sinxcosx+2(cos²x-sin²x)=1
3sin²+3sinxcosx+2cos²x-2sin²x=1
sin²+3sinxcosx+2cos²x=1
1+3sinxcosx+cos²x=1
3sinxcosx+cos²x=0
cosx(3sinx+cosx)=0
1. cosx=0
x=пn n=0,±1,±2...
п- это пи
2. 3sinx+cosx=0
3tgx+1=0
tgx=-1/3
x=-arctg(1/3)±2пn
A1=12
a14=-27
S14=a1+a14*14=12+(-27)*14=273
2. 2
Ответ:
x(1) = 1
x(2) = 2
x(3;4) = (3+-√33) / 2
Объяснение:
(x²-3x)(x²-3x-4)=12
x²-3x=t
t(t-4) = 12
t² - 4t -12 = 0
D = 16 + 48 = 64 = 8²
t(1) = (4-8)/2 = -4/2= -2
t(2) = (4+8) / 2 = 6
x²-3x=-2 x²-3x=6
x²-3x+2 = 0 x²-3x-6=0
D = 9-8 = 1 = 1² D = 9 + 24 = 33 = (√33)²
x(1) = (3+1) / 2 = 2 x(3;4) = (3+-√33) / 2
x(2) = (3-1) / 2 = 1