rzfansubs87

Answered

2022-07-07

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

Answer & Explanation

Zackery Harvey

Expert

2022-07-08Added 21 answers

Clearly, $\mathrm{cos}2x\mathrm{cos}x\ne 0$

Multiply both sides of the given equation

$\mathrm{cos}x\mathrm{sin}2x+3(1+\mathrm{cos}x)\mathrm{cos}2x=0$

$0=2\mathrm{sin}x{\mathrm{cos}}^{2}x+3(1+\mathrm{cos}x)\mathrm{cos}2x$

$=4\mathrm{sin}{\displaystyle \frac{x}{2}}\mathrm{cos}{\displaystyle \frac{x}{2}}{\mathrm{cos}}^{2}x+3\cdot 2{\mathrm{cos}}^{2}{\displaystyle \frac{x}{2}}\mathrm{cos}2x$

$=2\mathrm{cos}{\displaystyle \frac{x}{2}}(2\mathrm{sin}{\displaystyle \frac{x}{2}}{\mathrm{cos}}^{2}x+3\mathrm{cos}{\displaystyle \frac{x}{2}}\mathrm{cos}2x)$

If $\mathrm{cos}{\displaystyle \frac{x}{2}}=0,{\displaystyle \frac{x}{2}}=(2n+1){90}^{\circ}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=(2n+1){180}^{\circ}\equiv {180}^{\circ}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{360}^{\circ})$

Else $2\mathrm{sin}{\displaystyle \frac{x}{2}}{\mathrm{cos}}^{2}x+3\mathrm{cos}{\displaystyle \frac{x}{2}}\mathrm{cos}2x=0$

$-2\mathrm{tan}{\displaystyle \frac{x}{2}}={\displaystyle \frac{3(2{\mathrm{cos}}^{2}x-1)}{{\mathrm{cos}}^{2}x}}$

Use $\mathrm{cos}x={\displaystyle \frac{1-{\mathrm{tan}}^{2}{\displaystyle \frac{x}{2}}}{1+{\mathrm{tan}}^{2}{\displaystyle \frac{x}{2}}}}$ which unfortunately leaves us with a bi-quadratic equation in $\mathrm{tan}{\displaystyle \frac{x}{2}}$

Multiply both sides of the given equation

$\mathrm{cos}x\mathrm{sin}2x+3(1+\mathrm{cos}x)\mathrm{cos}2x=0$

$0=2\mathrm{sin}x{\mathrm{cos}}^{2}x+3(1+\mathrm{cos}x)\mathrm{cos}2x$

$=4\mathrm{sin}{\displaystyle \frac{x}{2}}\mathrm{cos}{\displaystyle \frac{x}{2}}{\mathrm{cos}}^{2}x+3\cdot 2{\mathrm{cos}}^{2}{\displaystyle \frac{x}{2}}\mathrm{cos}2x$

$=2\mathrm{cos}{\displaystyle \frac{x}{2}}(2\mathrm{sin}{\displaystyle \frac{x}{2}}{\mathrm{cos}}^{2}x+3\mathrm{cos}{\displaystyle \frac{x}{2}}\mathrm{cos}2x)$

If $\mathrm{cos}{\displaystyle \frac{x}{2}}=0,{\displaystyle \frac{x}{2}}=(2n+1){90}^{\circ}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=(2n+1){180}^{\circ}\equiv {180}^{\circ}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{360}^{\circ})$

Else $2\mathrm{sin}{\displaystyle \frac{x}{2}}{\mathrm{cos}}^{2}x+3\mathrm{cos}{\displaystyle \frac{x}{2}}\mathrm{cos}2x=0$

$-2\mathrm{tan}{\displaystyle \frac{x}{2}}={\displaystyle \frac{3(2{\mathrm{cos}}^{2}x-1)}{{\mathrm{cos}}^{2}x}}$

Use $\mathrm{cos}x={\displaystyle \frac{1-{\mathrm{tan}}^{2}{\displaystyle \frac{x}{2}}}{1+{\mathrm{tan}}^{2}{\displaystyle \frac{x}{2}}}}$ which unfortunately leaves us with a bi-quadratic equation in $\mathrm{tan}{\displaystyle \frac{x}{2}}$

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