Используя теорему Виета:
х²+px+q=0
A) x1+x2=-p
x1×x2=q
-p=(-1)+3=3-1
-p=2
q=(-1)×3
q=-3
x²-2x-3=0
Проверка:
D=(-(-2))²-4×1×(-3)=4+12=16
x1=(-(-2)-√16)/2×1=(2-4)/2=-2/2=-1
x2=(-(-2)+√16)/2×1=(2+4)/2=6/2=3
b) x1+x2=-p
-p=1/2+(-3/4)=1/2-3/4=0,5-0,75
-p=-0,25
-p=-1/4
x1×x2=q
q=1/2×(-3/4)=0,5×(-0,75)
q=-0,375=-(375/1000)=-3/8
x²+(1/4)x-(3/8)=0|×8
8x²+2x-3=0
Проверка:
D=(-2)²-4×8×(-3)=4+96=100
x1=(-2+√100)/2×8=(-2+10)/16=8/16=1/2
x2=(-2-√100)/2×8=(-2-10)/16=-12/16=-3/4
Ответ:
a) x²-2x-3=0
b) x²+(1/4)x-(3/8)=0 или 8x²+2x-3=0
-sin(30*a) - cos(60*a) = -sqrt(3) * sqrt(sin a)
Решение
8sin⁴<span>x + 10sin</span>²<span>x - 3 = 0
</span>sin²x = t, t ≥ 0
8t² + 10t - 3 = 0
D = 100 + 4*8*3 = 196
t₁ = (- 10 - 14)/16
t₁ = - 24/16
t₁ = - 1,5, не удовлетворяет условию t ≥ 0
t₂ = (- 10 + 14)/16
t₂ = 4/16
t₂ = 1/4
sin²x = 1/4
sinx = - 1/2
sinx = 1/2
1) sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n∈Z
x = (-1)^(n+1)*arcsin(1/2) + πn, n∈Z
x = (-1)^(n+1)*arcsin(1/2) + πn, n∈Z
x = (-1)^(n+1)*(π/6) + πn, n∈Z
2) sinx = 1/2
x = (-1)^k*arcsin(1/2) + πk, n∈Z
x = (-1)^k*(π/6) + πk, k∈Z