ответ: x ∈ {2*пи*k/3-5*пи/18, 2*пи*k/3-пи/18}, k ∈ Z
<span>y' = tg' x + ctg' x = 1/(cos^2 (x) ) + (- 1/(sin^2 (x) ) = 1/(cos^2 (x) ) - 1/(sin^2 (x) ) = 1/(sin^2 (x) * cos^2 (x) ) = [(1-cos(2x))/2 ]*[(1+cos(2x))/2] = (1-cos^2 (2x) )/4</span>
2 целых 18/31
-4 целых 4/31