Question Number 94877 by mathocean1 last updated on 21/May/20
![Solve for x in [0;2π]: sin(4x−(π/6))=(1/2)](Q94877.png)
$$\mathrm{Solve}\:\mathrm{for}\:{x}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\right]: \\ $$$$\mathrm{sin}\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathocean1 last updated on 21/May/20
![Please sir, is there only 2 solutions ∈ [0;2π] ?](Q94883.png)
$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{is}\:\mathrm{there}\:\mathrm{only}\:\mathrm{2}\:\mathrm{solutions}\: \\ $$$$\in\:\left[\mathrm{0};\mathrm{2}\pi\right]\:? \\ $$
Commented by mathocean1 last updated on 21/May/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 21/May/20
![4x−(π/6)=2kπ+(π/6) ⇒x=((kπ)/2)+(π/(12)) 4x−(π/6)=2kπ+π−(π/6) ⇒x=((kπ)/2)+(π/4) within [0, 2π]: x=(π/(12)), (π/4) x=((7π)/(12)), ((3π)/4) x=((13π)/(12)), ((5π)/4) x=((19π)/(12)), ((7π)/4)](Q94919.png)
$$\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{6}}\:\Rightarrow{x}=\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{12}} \\ $$$$\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}=\mathrm{2}{k}\pi+\pi−\frac{\pi}{\mathrm{6}}\:\Rightarrow{x}=\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}} \\ $$$${within}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]: \\ $$$${x}=\frac{\pi}{\mathrm{12}},\:\frac{\pi}{\mathrm{4}} \\ $$$${x}=\frac{\mathrm{7}\pi}{\mathrm{12}},\:\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$${x}=\frac{\mathrm{13}\pi}{\mathrm{12}},\:\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$${x}=\frac{\mathrm{19}\pi}{\mathrm{12}},\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$
Commented by ElOuafi last updated on 22/May/20
$${this}\:{is}\:{the}\:{correct}\:{answer}\:{sir}\:\left(\mathrm{8}\:{solutions}\right)! \\ $$