Question 1169241
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Let *[tex \Large u\ =\ tan(x)], then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4u^2\ +\ 2u\ -\ 3\ =\ 0]


I'll leave it to you to verify that the roots are *[tex \Large u\ =\ \frac{-1\,\pm\,\sqrt{13}}{4}]


So then *[tex \Large x\ \in\ -\frac{\pi}{2}\,<\,x\,<\,\frac{\pi}{2}\ =\ \arctan\(\frac{-1\,+\,\sqrt{13}}{4}\)\text{ or }\arctan\(\frac{-1\,-\,\sqrt{13}}{4}\)].


However, *[tex \Large \arctan\(\frac{-1\,-\,\sqrt{13}}{4}\)] is outside the domain interval specified for the problem, so to find the values of the angle for the negative root you need:  *[tex \Large \arctan\(\frac{-1\,-\,\sqrt{13}}{4}\)\ +\ \pi] and *[tex \Large \arctan\(\frac{-1\,-\,\sqrt{13}}{4}\)\ +\ 2\pi].


For the positive root of your quadratic you need: *[tex \Large \arctan\(\frac{-1\,+\,\sqrt{13}}{4}\)] and *[tex \Large \arctan\(\frac{-1\,+\,\sqrt{13}}{4}\)\ +\ \pi]

See the diagram


 *[illustration QuadraticInTanxForx.jpg].
												
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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