The general solution of $\mathrm{tan}(x)=\mathrm{tan}(2x-\pi /2)$

Hayley Mcclain
2022-11-20
Answered

The general solution of $\mathrm{tan}(x)=\mathrm{tan}(2x-\pi /2)$

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Savanna Smith

Answered 2022-11-21
Author has **17** answers

$\mathrm{tan}(2x-\frac{\pi}{2})=-\mathrm{tan}(\frac{\pi}{2}-2x)=-\frac{1}{\mathrm{tan}2x}$

$\mathrm{tan}x=-\frac{1}{\mathrm{tan}2x}\to \mathrm{tan}x\mathrm{tan}2x=-1$

$\mathrm{tan}x\frac{2\mathrm{tan}x}{1-{\mathrm{tan}}^{2}x}=-1\to 2{\mathrm{tan}}^{2}x=-1+{\mathrm{tan}}^{2}x\to {\mathrm{tan}}^{2}x=-1!$

$\mathrm{tan}x=-\frac{1}{\mathrm{tan}2x}\to \mathrm{tan}x\mathrm{tan}2x=-1$

$\mathrm{tan}x\frac{2\mathrm{tan}x}{1-{\mathrm{tan}}^{2}x}=-1\to 2{\mathrm{tan}}^{2}x=-1+{\mathrm{tan}}^{2}x\to {\mathrm{tan}}^{2}x=-1!$

Alberto Calhoun

Answered 2022-11-22
Author has **5** answers

If you set $t=\mathrm{tan}x$ and apply the duplication formula for the tangent, you have

$\mathrm{tan}{\textstyle (}2x-\frac{\pi}{2}{\textstyle )}=-\frac{1}{\mathrm{tan}2x}=\frac{{t}^{2}-1}{2t},$

so the equation becomes

$\frac{{t}^{2}-1}{2t}=t\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{t}^{2}-1=2{t}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{t}^{2}=-1,$

which has no real solution

$\mathrm{tan}{\textstyle (}2x-\frac{\pi}{2}{\textstyle )}=-\frac{1}{\mathrm{tan}2x}=\frac{{t}^{2}-1}{2t},$

so the equation becomes

$\frac{{t}^{2}-1}{2t}=t\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{t}^{2}-1=2{t}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{t}^{2}=-1,$

which has no real solution

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