1) 5cos2x -6cos²x +4 =0;
5cos2x -6(1+cos2x)/2 +4 =0 ;
cos2x = -1/2 ;
2x = (+/-) (π -π/3) +2π*k , k∈Z.
x = (+/-) π/3 + π*k , k∈Z.
ответ : (+/-) π/3 + π*k , k∈Z.
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2) 9cos2x +3cosx -1 =0;
9(2cos²x -1)+3cosx -1 =0;
18cos²x +3cosx -10 =0 ; *** t =cosx ; | t | ≤ 1 ***
18t² +3t -10 =0 ;
D =3² +4*18*10 =729 =27² ;
t ₁ = ( -3 -27)/36 = -5/6 ;
t ₂=( - 3 +27)/36 = 2/3 .
cosx = -5/6⇒x =(+/-)arccos(-5/6) +2π*k или x = (+/-)(π -arccos5/6)+2π*k ;k∈Z ;
cosx = 2/3 ⇒x =(+/-)arccos(2/3) +2π*k ;k∈Z .
ответ : (+/-)arccos(-5/6) +2π*k , (+/-)arccos(2/3) +2π*k ;k∈Z .
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3) 5sin2x -18cos²x+14 =0 ;
5*2sinxcosx -18cos²x +14(cos²x+sin²x) =0;
7sin²x +5sinxcosx - 2cos²x =0 ;
7tq²x +5tqx -2 =0 ;
7t² +5t -2 =0 ;
D =5² -4*7(-2) =81 =9² ;
t₁= (-5-9)/14=-1;
t₂ =( -5 +9)/14 = 2/7;
tqx = -1⇒ x = -π/4 +π*k , k∈Z.
tqx = 2/7⇒ x =arctq2/7 +π*k , k∈Z.
ответ : - π/4 +π*k , arctq2/7 +π*k , k∈Z .
4) 3cosx+11sinx+9 =0; ***√(3²+11²) cos(x -arctq11/3) = -9****
3(cos²x/2 -sin²x/2) +11*2sinx/2*cosx/2 + 9(sin²x/2 +cos²x/2) =0;
6sin²x/2 +22sinx/2*cosx/2 +12cos²x/2 =0 ;
3sin²x/2 +11sinx/2*cosx/2 +6cos²x/2 =0 ; ***cosx/2 ≠0 ***
3tq²x/2 +11tqx/2 +6 =0 ;
[ tqx/2 = -3 ;tqx/2 = -2/3.
{x= - 2arctq3+2π*k ; x = -2arctq2/3+2π *k ; k∈Z.
ответ : - 2arctq3+2π*k , x = -2arctq2/3+2π *k ; k∈Z.∈
Должно быть так, если нет - сорьки
y = -x^(3/2)/sqrt(2)
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y = x^(3/2)/sqrt(2)
Polynomial discriminant:
Δ_x = -108 y^4
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Integer roots:
x = 2, y = ± 2
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x = 8, y = ± 16
x = 18, y = ± 54
x = 0, y = 0
Properties as a function:Domain:
R^2
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Range:
R (all real numbers)
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Partial derivatives:Step-by-step solution
d/(dx)(2 y^2 - x^3) = -3 x^2
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d/(dy)(2 y^2 - x^3) = 4 y
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Indefinite integral:Step-by-step solution
integral(-x^3 + 2 y^2) dx = 2 x y^2 - x^4/4 + constant
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Definite integral over a disk of radius R:
integral integral_(x^2 + y^2<R^2)(2 y^2 - x^3) dx dy = (π R^4)/2
Definite integral over a square of edge length 2 L:
integral_(-L)^L integral_(-L)^L (-x^3 + 2 y^2) dy dx = (8 L^4)/3
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