sin(x²+x)=1/2
x²+x = (-1)ⁿarcsin(1/2) + πn, n∈Z;
x²+x = (-1)ⁿπ/6 + πn, n∈Z;
x² + x - (-1)ⁿπ/6 - πn = 0;
D = 1 + 4·((-1)ⁿπ/6 - πn);
D₁ = 1 + 4·(π/6 - 2πn); D₁ ≥ 0; 1 + 4·(π/6 - 2πn) ≥ 0; π/6 - 2πn ≥ -0,25| · 6;
π - 12πn ≥ -1,5; 12πn ≤ π + 1,5; n ≤ (π + 1,5)/(12π); n ≤ (2π + 3)/(24π);
n ≤ 1/12 + 3/(8π); n = 0; -1; -2;
x₁ = (-1 + √(1 + 4·(π/6 - 2πn)))/2; x₂ = (-1 - √(1 + 4·(π/6 - 2πn)))/2, n = 0; -1; -2;
D₂ = 1 + 4·(-π/6 - 2πn - π) = 1 + 4·(-7π/6 - 2πn) ; D₂ ≥ 0; 1 + 4·(-7π/6 - 2πn) ≥ 0;
-7π/6 - 2πn ≥ -1/4; 2πn ≤ 1/4 - 7π/6; n ≤ (1/4 - 7π/6)/(2π);
n ≤ 1/(8π) - 7/12; n = -1; -2; -3; ...
x₁ = (-1 + √(1 + 4·(-7π/6 - 2πn)))/2; x₂ = (-1 - √(1 + 4·(-7π/6 - 2πn)))/2, n = -1; -2; -3; ...
task/29383416
Решить уравнение sin(x²+x)=1/2
sin(x²+x)=1/2 ⇒
а)
x²+ x = c₁, где c₁ = π/6 + 2πk ; k ∈ ℤ * * * π/6 +π*(2k) =π/6 +π*n * **
x²+ x - c₁ = 0
если D = 1 + 4c₁ ≥ 0 ⇔ 1 + 2π/3 + 8πk ≥ 0 ⇔ k ≥ - 1/8π - 1 /12, т.е. k ∈ ℤ₊
* * * ℤ₊ ={ 0 ; 1 ; 2: } * * *
x₁ , ₂ = (- 1 ±√( 1 + 2π/3 + 8πk) ) /2 , k ∈ ℤ₊
б)
x²+ x = c₂ , где c₂ = 5π/6 +2πk ; k ∈ ℤ
* * * π - π/6 +2πk = - π/6+π(2к+1) = - π/6+πn , nнечетное целое число * * *
x²+ x - c₂=0 ₂ ≥ 0 ⇔ 1 + 10π/3 + 8πk ≥ 0 ⇔ k ≥ - 1/8π - 5/12, т.е. k ∈ ℤ₊
x₃ , ₄ = (- 1 ±√( 1 + 10π/3 + 8πk) ) /2 , k ∈ ℤ₊
При отрицательных целых k уравнение не имеет решения