Уравнение имеет решение, если sinx - cosx ≥ 0; √2sin(x - π/4) ≥ 0; sin(x - π/4) ≥ 0; 2πn ≤ x - π/4 ≤ π + 2πn, n∈Z; π/4 + 2πn ≤ x ≤ 5π/4 + 2πn, n∈Z.
1,5 + cos²x = (sinx - cosx)²;
1,5 + cos²x = sin²x + cos²x - 2sinxcosx;
1,5 + cos²x = 1 - 2sinxcosx;
cos²x + 2sinxcosx +0,5 = 0;
2cos²x + 4sinxcosx + 1 = 0| : sin²x;
2ctg²x + 4ctgx + 1/sin²x = 0;
2ctg²x + 4ctgx + 1 + ctg²x = 0;
3ctg²x + 4ctgx + 1 = 0;
Замена: ctgx = t/3
t² + 4t + 3 = 0;
t₁ = -1; t₂ = -3
Обратная замена:
ctgx = -1 или ctgx = -1/3
x₁ = 3π/4 + πn, n∈Z; x₂ = arcсtg(-1/3) + πn, n∈Z.
Данное уравнение удовлетворяют значения х₁ = 3π/4 + 2πn, n∈Z; x₂ = arcсtg(-1/3) + 2πn = -arcсtg(1/3) + π(2n+1), n∈Z.
ответ: 3π/4 + 2πn, n∈Z; -arcctg(1/3) + π(2n + 1), n∈Z.
Замена: cos(x/3) = t, -1 ≤ t ≤ 1
2t^2 + 3t - 2 = 0
D = 9 + 4*2*2 = 9 + 16 = 25
t1 = (-3 + 5)/4 = 2/4 = 1/2
t2 = (-3 - 5)/4 = -8/4 = -2 < -1 - посторонний корень
cos(x/3) = 1/2
(x/3) = +-π/3 + 2πk
x = +- π + 6πk
2) Замена: cos(x/2) = t, -1≤ t ≤ 1
2t^2 + √3*t = 0, t*(2t + √3) = 0
t1 = 0
t2 = -√3/2
cos(x/2) = 0, (x/2) = 2πk, x = 4πk
cos(x/2) = -√3/2, x/2 = +-5π/6 + 2πk, x = +-5π/3 + 4πk