All x satisfying the inequality (cot−1x)2−7(cot−1x)+10>0, lie in the interval :
Solutions · 1
Official solution
(cot−1x−5)(cot−1x−2)>0⇒cot−1x∈(−∞,2)∪(5,∞).
Since cot−1x∈(0,π) always, and π<5, the branch (5,∞) is impossible. So cot−1x∈(0,2).
Since cot−1 is decreasing, this corresponds to x∈(cot2,∞).