Думаю,как-то так.............
{x+3y=17
{x²+y²=29
{x=17-3y
{(17-3y)²+y²=29
289+9y²-102y+y²=29
10y²-102y+260=0
5y²-51y+130=0
D=1
y₁=(51+1)/10=5,2
y₂=(51-1)/10=5
x₁=17-3*5,2=1,4
x₂=17-3*5=2
Ответ: (1,4; 5,2), (2; 5)
Касательные параллельны оси абсцисс в точках экстремумов.
1) y' = 12x^3 - 84x^2 - 12x + 84 = 12(x-7)(x^2 - 1) = 12(x-7)(x-1)(x+1) = 0
x1 = -1; y(-1) = 3 + 28 - 6 - 84 + 1 = -58
x2 = 1; y(1) = 3 - 28 - 6 + 84 + 1 = 54
x3 = 7; y(7) = 3*2401 - 28*343 - 6*49 + 84*7 + 1 = -2106
2) y' = -2sin 2x + 5sin x = -4sin x*cos x + 5sin x = sin x*(5 - 4cos x) = 0
sin x = 0; x1 = 2pi*k; y(x1) = cos(4pi*k) - 5cos(2pi*k) = 1 - 5*1 = -4
x2 = pi + 2pi*k; y(x2) = cos(2pi+4pi*k) - 5(pi+2pi*k) = 1 - 5(-1) = 6
5 - 4cos x = 0; cos x = 5/4 > 1 - решений нет.
3) y' = (x - 4)^3 + x*3(x - 4)^2 = (x - 4)^2*(x - 4 + 3x) = (x - 4)^2*(4x - 4) = 0
x1 = x2 = 4; y(4) = 0
x3 = 1; y(1) = 1*(1 - 4)^3 = 1(-3)^3 = -27
sin4a/cos6x=cos2a/sin4a
sin^2(4a)=cos6a*cos2a=0.5(cos8a+cos4a)
пусть 4a=y
2sin^2y=cos2y+cosy
2(1-cos^2y)=cos^2y-sin^2y+cosy
2-2cos^2y-cos^2y+sin^2y-cosy=0
2-3cos^2y+(1-cos^2y)-cosy=0
3-4cos^2y-cosy=0
пусть cosy=t
3-4t^2-t=0
D=1^2-4(-4)*3=49
t1=(1+7)/(-8)=-1; cos4x=-1; 4x=pi+2pil; x=pi/4+pik/2
t2=(1-7)/(-8)=3/4; cos4x=3/4; 4x=+-arcc0s(3/4)+2pik; x=+-0.25 arccos(3/4)+pik/2