Menu Close

3ix-pi-2-ix-5pi-6-0-x-

Tinku Tara 0 Comments
Pin




Question Number 222521 by hu last updated on 29/Jun/25
−3ix+(π/2)+((ix−(√(5π)))/6)=0 x=?
$$−\mathrm{3}{ix}+\frac{\pi}{\mathrm{2}}+\frac{{ix}−\sqrt{\mathrm{5}\pi}}{\mathrm{6}}=\mathrm{0} \\ $$$${x}=?\: \\ $$
Commented by Rasheed.Sindhi last updated on 29/Jun/25
(√(5π)) or (√5) π ?
$$\sqrt{\mathrm{5}\pi}\:{or}\:\sqrt{\mathrm{5}}\:\pi\:? \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jun/25
−3ix+(π/2)+((ix−(√5) π)/6)=0 −3ix+(i/6)x+(π/2)−(((√5) π)/6)=0 x(−3+(1/6))i=((((√5) )/6)−(1/2))π x=((((((√5) )/6)−(1/2))π)/((−3+(1/6))i)) =((((((√5) −3)/6))πi)/((((17)/6)))) =((((√5) −3)πi)/(17))
$$−\mathrm{3}{ix}+\frac{\pi}{\mathrm{2}}+\frac{{ix}−\sqrt{\mathrm{5}}\:\pi}{\mathrm{6}}=\mathrm{0} \\ $$$$−\mathrm{3}{ix}+\frac{{i}}{\mathrm{6}}{x}+\frac{\pi}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}\:\pi}{\mathrm{6}}=\mathrm{0} \\ $$$${x}\left(−\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}}\right){i}=\left(\frac{\sqrt{\mathrm{5}}\:}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi \\ $$$${x}=\frac{\left(\frac{\sqrt{\mathrm{5}}\:}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi}{\left(−\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}}\right){i}} \\ $$$$\:\:=\frac{\left(\frac{\sqrt{\mathrm{5}}\:−\mathrm{3}}{\mathrm{6}}\right)\pi{i}}{\left(\frac{\mathrm{17}}{\mathrm{6}}\right)} \\ $$$$\:\:\:=\frac{\left(\sqrt{\mathrm{5}}\:−\mathrm{3}\right)\pi{i}}{\mathrm{17}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *