fythynwyrk0

2022-07-13

Solve $\frac{2}{\mathrm{sin}x\mathrm{cos}x}=1+3\mathrm{tan}x$

isscacabby17

Expert

Note that $\mathrm{tan}x+\mathrm{cot}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}+\frac{\mathrm{cos}x}{\mathrm{sin}x}=\frac{1}{\mathrm{sin}x\mathrm{cos}x}$ (this is something I've filed away from seeing it in a lot of "verify this trig identity" type problems) .
So your equation can be written $2\left(\mathrm{tan}x+\mathrm{cot}x\right)=1+3\mathrm{tan}x$, or
$2\left(\mathrm{tan}x+\frac{1}{\mathrm{tan}x}\right)=1+3\mathrm{tan}x.$
Now that everything is in terms of tangent, you should be able to get it into the form of an equation that's quadratic in $\mathrm{tan}x$ (i.e., has multiples of ${\mathrm{tan}}^{2}x,\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}x,$, and a number) and solve; the values of tanx are nice integers, although you will need inverse trig to get one of the x-values.

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