sgrt (cos^2x-sin^2x) * (tg2x-1) = 0

sgrt (cos^2x-sin^2x) * (tg2x-1) = 0

sgrt (cos^2x-sin^2x) = 0 или (tg2x-1) = 0

возведем в квадрат: tg2x = 1

(cos^2x-sin^2x) = 0 2 х = arctg (1) + πk, k∈Z

1-sin^2x-sin^2x=0 2 х = π/4 + πk, k∈Z

1-2sin^2x=0 х = π/8 + πk/2, k∈Z

2sin^2x=1

sin^2x=1/2

sinx=±√2/2

х = (-1) ^k * arcsin (±√2/2) + πk, k∈Z

х = (-1) ^k * (±π/4) + πk, k∈Z

Ответ: х = (-1) ^k * (±π/4) + πk, k∈Z х = π/8 + πk/2, k∈Z