Sin^2x-cos^2=Корень2/2
Применим формулу: cos2α = cos∧2x - sin∧2x; -( cos∧2x - sin∧x) = - cos2α - cos(2α) = √2/2 cos(2α) = - √2/2 2α = (+ -) arccos (√2/2) + 2πn, n∈Z 2α =( + -) (π - π/4) + 2πn,n∈Z 2α = (+ -) (3π/4) + 2πn. n∈Z α =( + -) (3π/8) + πn, n∈Z