rzfansubs87

2022-07-07

Trigonometry equation $\mathrm{tan}2x+3\mathrm{sec}x+3=0$ for $\mathrm{tan}2x+3\mathrm{sec}x+3=0$

Zackery Harvey

Expert

Clearly, $\mathrm{cos}2x\mathrm{cos}x\ne 0$
Multiply both sides of the given equation
$\mathrm{cos}x\mathrm{sin}2x+3\left(1+\mathrm{cos}x\right)\mathrm{cos}2x=0$
$0=2\mathrm{sin}x{\mathrm{cos}}^{2}x+3\left(1+\mathrm{cos}x\right)\mathrm{cos}2x$
$=4\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}{\mathrm{cos}}^{2}x+3\cdot 2{\mathrm{cos}}^{2}\frac{x}{2}\mathrm{cos}2x$
$=2\mathrm{cos}\frac{x}{2}\left(2\mathrm{sin}\frac{x}{2}{\mathrm{cos}}^{2}x+3\mathrm{cos}\frac{x}{2}\mathrm{cos}2x\right)$
If $\mathrm{cos}\frac{x}{2}=0,\frac{x}{2}=\left(2n+1\right){90}^{\circ }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=\left(2n+1\right){180}^{\circ }\equiv {180}^{\circ }\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{360}^{\circ }\right)$
Else $2\mathrm{sin}\frac{x}{2}{\mathrm{cos}}^{2}x+3\mathrm{cos}\frac{x}{2}\mathrm{cos}2x=0$
$-2\mathrm{tan}\frac{x}{2}=\frac{3\left(2{\mathrm{cos}}^{2}x-1\right)}{{\mathrm{cos}}^{2}x}$
Use $\mathrm{cos}x=\frac{1-{\mathrm{tan}}^{2}\frac{x}{2}}{1+{\mathrm{tan}}^{2}\frac{x}{2}}$ which unfortunately leaves us with a bi-quadratic equation in $\mathrm{tan}\frac{x}{2}$

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