Решите уравнение

(2 cos^2x-2 cosx+√2-1) / (cos^2x-sin^2⁡x) = 0

Решите уравнение

(2 cos^2x-2 cosx+√2-1) / (cos^2x-sin^2⁡x) = 0

одз:

(cos^2x-sin^2⁡x) ≠0 или tg²x≠1 tgx≠1 tgx≠ - 1

cos2x≠0 2x≠ π/2+πn, n∈Z x≠π/4+πn/2.

(2 cos^2x-2 cosx+√2-1) / (cos^2x-sin^2⁡x) = 0

(2 cos^2x-2 cosx+√2-1) = 0 t=cosx ⇒2t²-2t+√2-1

t1=1-√2 / 2 t2=1/√2

1) cosx=1-√2 / 2

x1=arccos (1-√2 / 2) + 2πn, n∈Z

x2 = - arccos (1-√2 / 2) + 2πn, n∈Z

2) cosx=1/√2 ∉одз