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Question Number 122337 by benjo_mathlover last updated on 15/Nov/20

Find domain of function r(x) = arc cos ((3/(4+2sin x)))

$$\:{Find}\:{domain}\:{of}\:{function} \\ $$$$\:{r}\left({x}\right)\:=\:\mathrm{arc}\:\mathrm{cos}\:\:\left(\frac{\mathrm{3}}{\mathrm{4}+\mathrm{2sin}\:{x}}\right) \\ $$

Commented by liberty last updated on 16/Nov/20

The function defined for values of x for which −1≤ (3/(4+2sin x)) ≤ 1 . Since 4+2sin x > 0 at any x ,the problem is reduced to solving inequality (3/(4+2sin x)) ≤ 1 , whence 3 ≤ 4+2sin x i.e sin x ≥ −(1/2) ⇒ −(π/6)+2kπ ≤ x≤ ((7π)/6)+2kπ for k ε Z. ▲

$$\mathrm{The}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{for}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x} \\ $$$$\mathrm{for}\:\mathrm{which}\:−\mathrm{1}\leqslant\:\frac{\mathrm{3}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{x}}\:\leqslant\:\mathrm{1}\:. \\ $$$$\mathrm{Since}\:\mathrm{4}+\mathrm{2sin}\:\mathrm{x}\:>\:\mathrm{0}\:\mathrm{at}\:\mathrm{any}\:\mathrm{x}\:,\mathrm{the}\:\mathrm{problem} \\ $$$$\mathrm{is}\:\mathrm{reduced}\:\mathrm{to}\:\mathrm{solving}\:\mathrm{inequality}\: \\ $$$$\:\frac{\mathrm{3}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{x}}\:\leqslant\:\mathrm{1}\:,\:\mathrm{whence}\:\mathrm{3}\:\leqslant\:\mathrm{4}+\mathrm{2sin}\:\mathrm{x} \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{sin}\:\mathrm{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:−\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\:\leqslant\:\mathrm{x}\leqslant\:\frac{\mathrm{7}\pi}{\mathrm{6}}+\mathrm{2k}\pi \\ $$$$\:\mathrm{for}\:\mathrm{k}\:\epsilon\:\mathbb{Z}.\:\blacktriangle \\ $$

Commented by benjo_mathlover last updated on 16/Nov/20

Commented by benjo_mathlover last updated on 16/Nov/20

correct..thanks

$${correct}..{thanks} \\ $$

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