Correct answer: (tan(1/2), cot 1)
First inequality: with u = arctan x, 4u² - 8u + 3 < 0 factors as (2u - 1)(2u - 3) < 0, so 1/2 < u < 3/2, i.e. 1/2 < arctan x < 3/2. Since arctan x < pi/2 ~ 1.571, the upper bound 3/2 is allowed: x in (tan(1/2), tan(3/2)). Second: with v = arccot x, 4v - v² - 3 >= 0 => v² - 4v + 3 <= 0 => (v-1)(v-3) <= 0 => 1 <= v <= 3. Since arccot x in (0, pi), this gives 1 <= arccot x <= 3, i.e. x in [cot 3, cot 1] (note arccot decreasing). Intersecting the two solution sets in x yields x in (tan(1/2), cot 1).