Question Number 87737 by M±th+et£s last updated on 05/Apr/20
![solve sin((π/([(([x])/4)])))=(1/2)](Q87737.png)
$${solve} \\ $$$${sin}\left(\frac{\pi}{\left[\frac{\left[{x}\right]}{\mathrm{4}}\right]}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by mahdi last updated on 06/Apr/20
![u=[(([x])/4)]⇒−1≤(1/u)≤1⇒−π≤(π/u)≤π {u≠0⇒x∉[0,4)} sin((π/u))=(1/2)⇒ { (((π/u)=(π/6)+2kπ)),(((π/u)=((5π)/6)+2kπ)) :} −π≤(π/u)≤π ⇒(π/u)=(π/6) ∨ ((5π)/6) u=6 ∨ (6/5)⇒^(u∈Z) u=6 [(([x])/4)]=6⇒24≤x<28](Q87746.png)
$$\mathrm{u}=\left[\frac{\left[\mathrm{x}\right]}{\mathrm{4}}\right]\Rightarrow−\mathrm{1}\leqslant\frac{\mathrm{1}}{\mathrm{u}}\leqslant\mathrm{1}\Rightarrow−\pi\leqslant\frac{\pi}{\mathrm{u}}\leqslant\pi\:\:\:\left\{\mathrm{u}\neq\mathrm{0}\Rightarrow\mathrm{x}\notin\left[\mathrm{0},\mathrm{4}\right)\right\} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{u}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\begin{cases}{\frac{\pi}{\mathrm{u}}=\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi}\\{\frac{\pi}{\mathrm{u}}=\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2k}\pi}\end{cases} \\ $$$$−\pi\leqslant\frac{\pi}{\mathrm{u}}\leqslant\pi\:\:\Rightarrow\frac{\pi}{\mathrm{u}}=\frac{\pi}{\mathrm{6}}\:\vee\:\frac{\mathrm{5}\pi}{\mathrm{6}}\: \\ $$$$\mathrm{u}=\mathrm{6}\:\vee\:\frac{\mathrm{6}}{\mathrm{5}}\overset{\mathrm{u}\in\mathrm{Z}} {\Rightarrow}\mathrm{u}=\mathrm{6} \\ $$$$\left[\frac{\left[\mathrm{x}\right]}{\mathrm{4}}\right]=\mathrm{6}\Rightarrow\mathrm{24}\leqslant\mathrm{x}<\mathrm{28} \\ $$
Commented by M±th+et£s last updated on 05/Apr/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
$${sir}\:{if}\:−\mathrm{24}\leqslant{x}<−\mathrm{20}\:\:\:\:\rightarrow\:\rightarrow\:{x}=−\mathrm{0}.\mathrm{5}\:{not}\:\mathrm{1}\: \\ $$$${so}\:{S}\left\{\mathrm{24}\leqslant{x}<\mathrm{8}\right\} \\ $$
Commented by mahdi last updated on 06/Apr/20
![what?how[−24≤x<−20⇒x=−0.5]?](Q87774.png)
$$\mathrm{what}?\mathrm{how}\left[−\mathrm{24}\leqslant\mathrm{x}<−\mathrm{20}\Rightarrow\mathrm{x}=−\mathrm{0}.\mathrm{5}\right]? \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
$${sorry}\:{i}\:{mean}\:{ans}=−\mathrm{0}.\mathrm{5}\:{if}\:−\mathrm{24}\leqslant{x}<\mathrm{20} \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
Commented by mahdi last updated on 06/Apr/20
![ok ser,i sorry. { (((π/6)+2kπ⇒...,((−11π)/6),(π/6),((13π)/6),...)),((((5π)/6)+2kπ⇒...,((−7π)/6),((5π)/6),((17π)/6),...)) :} only: (π/6) and ((5π)/6) ∈[−1,1]](Q87812.png)
$$\mathrm{ok}\:\mathrm{ser},\mathrm{i}\:\mathrm{sorry}.\begin{cases}{\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\Rightarrow...,\frac{−\mathrm{11}\pi}{\mathrm{6}},\frac{\pi}{\mathrm{6}},\frac{\mathrm{13}\pi}{\mathrm{6}},...}\\{\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2k}\pi\Rightarrow...,\frac{−\mathrm{7}\pi}{\mathrm{6}},\frac{\mathrm{5}\pi}{\mathrm{6}},\frac{\mathrm{17}\pi}{\mathrm{6}},...}\end{cases} \\ $$$$\mathrm{only}:\:\:\frac{\pi}{\mathrm{6}}\:\mathrm{and}\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
$${thank}\:{you}\:{sir}\: \\ $$