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Question Number 15301 by Tinkutara last updated on 09/Jun/17
With the help of graph, find the  solution set of inequation tan x > −(√3) .
$$\mathrm{With}\:\mathrm{the}\:\mathrm{help}\:\mathrm{of}\:\mathrm{graph},\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequation}\:\mathrm{tan}\:{x}\:>\:−\sqrt{\mathrm{3}}\:. \\ $$
Answered by mrW1 last updated on 09/Jun/17
x∈(nπ−(π/3), nπ+(π/2)) ∧ n∈Z
$$\mathrm{x}\in\left(\mathrm{n}\pi−\frac{\pi}{\mathrm{3}},\:\mathrm{n}\pi+\frac{\pi}{\mathrm{2}}\right)\:\wedge\:\mathrm{n}\in\mathbb{Z} \\ $$
Commented by Tinkutara last updated on 10/Jun/17
But answer is [nπ − (π/3) < x < nπ + (π/2)]  ∪ {nπ + ((2π)/3) < x < nπ + ((3π)/2)}  Your first set is correct but how to get  the 2^(nd)  set in union?
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\left[{n}\pi\:−\:\frac{\pi}{\mathrm{3}}\:<\:{x}\:<\:{n}\pi\:+\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\cup\:\left\{{n}\pi\:+\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:<\:{x}\:<\:{n}\pi\:+\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right\} \\ $$$$\mathrm{Your}\:\mathrm{first}\:\mathrm{set}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{set}\:\mathrm{in}\:\mathrm{union}? \\ $$
Commented by mrW1 last updated on 10/Jun/17
the second set is the same as the first.   {nπ + ((2π)/3) < x < nπ + ((3π)/2)}  ≡{(n+1)π−(π/3)<x<(n+1)π+(π/2)}  ≡{mπ−(π/3)<x<mπ+(π/2)}  since n ∈ Z, in set 1 is set 2 included.
$$\mathrm{the}\:\mathrm{second}\:\mathrm{set}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{the}\:\mathrm{first}. \\ $$$$\:\left\{{n}\pi\:+\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:<\:{x}\:<\:{n}\pi\:+\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right\} \\ $$$$\equiv\left\{\left(\mathrm{n}+\mathrm{1}\right)\pi−\frac{\pi}{\mathrm{3}}<\mathrm{x}<\left(\mathrm{n}+\mathrm{1}\right)\pi+\frac{\pi}{\mathrm{2}}\right\} \\ $$$$\equiv\left\{\mathrm{m}\pi−\frac{\pi}{\mathrm{3}}<\mathrm{x}<\mathrm{m}\pi+\frac{\pi}{\mathrm{2}}\right\} \\ $$$$\mathrm{since}\:\mathrm{n}\:\in\:\mathbb{Z},\:\mathrm{in}\:\mathrm{set}\:\mathrm{1}\:\mathrm{is}\:\mathrm{set}\:\mathrm{2}\:\mathrm{included}. \\ $$
Commented by Tinkutara last updated on 10/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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