Haialarmz6

2022-01-22

Find a solution to the equation
$\frac{1}{{\mathrm{sin}}^{2}\left(2x\right)}+\mathrm{tan}\left(x\right)-\frac{1}{\mathrm{tan}\left(x\right)}=2$
Assuming $\mathrm{sin}\left(x\right)\ne 0$ and $\mathrm{cos}\left(x\right)\ne 0$, I simplified the above to
$4{\mathrm{sin}}^{2}\left(x\right)-1-4\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=0$
which is equivalent to
$\mathrm{cos}\left(2x\right)+\mathrm{sin}\left(2x\right)=\frac{1}{2}$
I do not know what to do next. I would be grateful for any hints.

Jordyn Horne

We have that by linear combination of sine and cosine
$\mathrm{cos}\left(2x\right)+\mathrm{sin}\left(2x\right)=\sqrt{2}\mathrm{sin}\left(2x+\frac{\pi }{4}\right)=\frac{1}{2}$

immablondevl

Use $\mathrm{tan}2A=\frac{2\mathrm{tan}A}{1-{\mathrm{tan}}^{2}A}$
$\frac{1}{\mathrm{tan}x}-\mathrm{tan}x=\dots =2\mathrm{cot}2x$
$\frac{1}{{\mathrm{sin}}^{2}2x}=1+{\mathrm{cot}}^{2}2x$
On simplification
$1=\frac{2\mathrm{tan}2x}{1-{\mathrm{tan}}^{2}2x}=\mathrm{tan}4x$

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