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Solve-in-0-2pi-2sin-4x-pi-6-1-Please-sirs-




Question Number 94910 by mathocean1 last updated on 21/May/20
Solve in [0;2π]  2sin(4x−(π/6))≥1  Please sirs...
$$\mathrm{Solve}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\right] \\ $$$$\mathrm{2sin}\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right)\geqslant\mathrm{1} \\ $$$$\mathrm{Please}\:\mathrm{sirs}… \\ $$
Answered by mr W last updated on 21/May/20
sin (4x−(π/6))≥(1/2)  2kπ+(π/6)≤4x−(π/6)≤2kπ+π−(π/6)  ⇒((kπ)/2)+(π/(12))≤x≤((kπ)/2)+(π/4) with k∈Z  within [0, 2π]:  (π/(12))≤x≤(π/4)  ((7π)/(12))≤x≤((3π)/4)  ((13π)/(12))≤x≤((5π)/4)  ((9π)/(12))≤x≤((7π)/4)
$$\mathrm{sin}\:\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right)\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{6}}\leqslant\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\leqslant\mathrm{2}{k}\pi+\pi−\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\:{with}\:{k}\in\mathbb{Z} \\ $$$${within}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]: \\ $$$$\frac{\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{7}\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{13}\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{9}\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$

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