Question Number 94910 by mathocean1 last updated on 21/May/20
![Solve in [0;2π] 2sin(4x−(π/6))≥1 Please sirs...](Q94910.png)
$$\mathrm{Solve}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\right] \\ $$$$\mathrm{2sin}\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right)\geqslant\mathrm{1} \\ $$$$\mathrm{Please}\:\mathrm{sirs}... \\ $$
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)](Q94918.png)
$$\mathrm{sin}\:\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right)\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{6}}\leqslant\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\leqslant\mathrm{2}{k}\pi+\pi−\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\:{with}\:{k}\in\mathbb{Z} \\ $$$${within}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]: \\ $$$$\frac{\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{7}\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{13}\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{9}\pi}{\mathrm{12}}\leqslant{x}\leqslant\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$