Решите уравнение: а)y³-6y=0 b)6x⁴+3,6x²=0 c)x³-0,1x=0,3x² d)x³+3x=3,5x² e)9x³-18x²-x+2=0 f)y⁴-y³-16y²+16y=0 - id4038720210320 от vhjklv 20.03.2021 06:00

Question

Алгебра: Решите уравнение: а)y³-6y=0 b)6x⁴+3,6x²=0 c)x³-0,1x=0,3x² d)x³+3x=3,5x² e)9x³-18x²-x+2=0 f)y⁴-y³-16y²+16y=0 g)p³-p²=p-1 h)x⁴-x²=3x³-3x

vhjklv · Accepted Answer

Cos (π/2 - t) - Sin (π+t) = √2
sint + sint = √2
2sint = √2
sint = √2/2
t = (-1)^(n)*arcsin(√2/2) + πn, n∈Z
t = (-1)^(n)*(π/4) + πn, n∈Z

2) Sin x/3 = -1/2
x = (-1)^(n)*arcsin(-1/2) + πk, n∈Z
x/3 = (-1)^(n+1)*arcsin(1/2) + πk, k∈Z
x/3 = (-1)^(n+1)*(π/6) + πk, k∈Z
x = (-1)^(n+1)*(3π/6) + 3πk, k∈Z
x = (-1)^(n+1)*(π/2) + 3πk, k∈Z

3) 5 Cos^2 x + 6 Sin x - 6 = 0
5*(1 - sin^2x) + 6sinx - 6 = 0
5 - 5*(sin^2x) + 6sinx - 6 = 0
5*(sin^2x) - 6sinx + 1 = 0
D = 36 - 4*5*1 = 16
a) sinx = (6 - 4)/10
sinx = 1/5
x = (-1)^(n)*arcsin(1/5) + πn, n∈Z
б) sinx = (6 + 4)/10
sinx = 1
x = π/2 + 2πk, k∈Z

Решите уравнение: а)y³-6y=0 b)6x⁴+3,6x²=0 c)x³-0,1x=0,3x² d)x³+3x=3,5x² e)9x³-18x²-x+2=0 f)y⁴-y³-16y²+16y=0 g)p³-p²=p-1 h)x⁴-x²=3x³-3x

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