1/((π/2)+πk) < 1+x2 ≤ 1/((π/4)+ πk ), k ∈ Z
(2/(π+2πk)) –1 < x2 ≤ (4/(π+4πk)) – 1, k ∈ Z
(2–π–2πk)/(π+2πk) < x2 ≤ (4–π–4πk)/(π+4πk)) , k ∈ Z
(4–π–4πk)/(π+4πk)) > 0 при k=0
(2–π–2πk)/(π+2πk) < 0 при любом k
x2 > (2–π–2πk)/(π+2πk) при любом х
Решаем неравенство:
x2 ≤ (4–π)/π
O т в е т. – √(4–π)/π ≤ x ≤ √(4–π)/π