Ответы на вопрос:
Пишу сразу ответ 1) (3-b)×(3+b)-2b×-b))=(3-b)×(3+b)+2b×(3-b)=(3-b)×(3+b+2b)=(3-b)×(3+3b)=(3-b)×3(1+b)=3(3-b)×(1+b) 2) -5y²-15y+2y-8=-5y²-13y-8 3) 6(x-9)-6x=6x-54-6x=-54
1) sinx+1/2 = 0 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)*(π/6) + πn, n∈z 2) 2sin^2x - cos2x=1 2sin^2x - (1 - 2 sin^2x) = 1 4sin^2x - 2 = 0 sin^2x = 2/4 a) sinx = - 1/2 x = (-1)^n*arcsin(-1/2) + πn, n∈z x = (-1)^(n+1)*arcsin(1/2) + πn, n∈z x1 = (-1)^(n+1)*(π/6) + πn, n∈z b) sinx = 1/2 x = (-1)^(n)*arcsin(1/2) + πk, n∈z x2 = (-1)^(n)*(π/6) + πk, k∈z 3) ctg^2x=3 a) ctgx = - √3 x1 = 5π/6 + πn, n∈z b) ctgx = √3 x2 = π/6 + πk, k∈z 4) sin^2x - 4sinx = 5 sin^2x - 4sinx - 5 = 0 sinx = t t^2 - 4t - 5 = 0 d = 16 + 4*1*5 = 36 t1 = (4 - 6)/2 t1 = - 1 t2 = (4 + 6)/2 t2 = 5 a) sinx = - 1 x = - π/2 + 2πn, n∈z sinx = 5 не удовлетворяет условию: i sinx i ≤ 1 5) 2sin2x*cos2x - 1= 0 sin(4x) - 1 = 0 sin(4x) = 1 4x = π/2 + 2πn, n∈z x = π/8 + πn/2, n∈z 6) tg(x/2) = √3 x/2 = arctg(√3) + πn, n∈z x/2 = π/3 + πn, n∈z x = 2π/3 + 2πn, n∈z 7) cos^2x-sin^2x=-1/2 cos(2x) = -1/2 2x = (+ -)*arccos(-1/2) + 2πn, n∈z 2x = (+ -)*(π - arccos(1/2)) + 2πn, n∈z 2x = (+ -)*(π - π/3) + 2πn, n∈z 2x = (+ -)*(2π/3) + 2πn, n∈z x = (+ -)*(π/3) + πn, n∈z 8) ctg(n/2 x-n) = 1 не понятен
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