Ashlyn Krause

2022-07-27

Solve the equation on the interval $\left[0,2\pi \right)$
$\mathrm{tan}x+\mathrm{sec}x=1$

Minbutastc

Expert

Given equation is $\mathrm{tan}x+\mathrm{sec}x=1$
$⇒\frac{\mathrm{sin}x}{\mathrm{cos}x}+\frac{1}{\mathrm{cos}x}=1$
$⇒\mathrm{sin}x+1=\mathrm{cos}x$
$⇒\mathrm{sin}x-\mathrm{cos}x=-1$
squaring on both sides, we get
$\left(\mathrm{sin}x-\mathrm{cos}x{\right)}^{2}=\left(-1{\right)}^{2}$
$⇒{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x-2\mathrm{sin}x\mathrm{cos}x=1$
$⇒1-2\mathrm{sin}x\mathrm{cos}x=1$
$⇒2\mathrm{sin}x\mathrm{cos}x=0$
$⇒\mathrm{sin}2x=0$
$⇒2x={\mathrm{sin}}^{-1}\left(0\right)$
$⇒2x=0,\pi ,2\pi$
$⇒x=0,\frac{\pi }{2},\pi$

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