Question 497224
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan^2\varphi\ +\ 3\sec\varphi\ +\ 3\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^2\varphi}{\cos^2\varphi}\ +\ \frac{3}{\cos\varphi}\ +\ 3\ =\ 0]


Multiply by *[tex \Large \cos^2\varphi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\varphi\ +\ 3\cos\varphi\ +\ 3\cos^2\varphi\ =\ 0]


Substitute from the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \1\ -\ \cos^2\varphi\ +\ 3\cos\varphi\ +\ 3\cos^2\varphi\ =\ 0]


Collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2\varphi\ +\ 3\cos\varphi\ +\ 1\ =\ 0]


Let *[tex \Large u\ =\ \cos\varphi] and substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2u^2\ +\ 3u\ +\ 1\ =\ 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2u\ +\ 1)(u\ +\ 1)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -\frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -1]


Substitute back:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\varphi\ =\ -\frac{1}{2}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\varphi\ =\ -1]


Use the Unit Circle:


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


to determine that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ =\ \cos^{-1}\,\left(-\frac{1}{2}\right)\ =\ \frac{2\pi}{3}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ =\ \cos^{-1}\,\left(-\frac{1}{2}\right)\ =\ \frac{4\pi}{3}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ =\ \cos^{-1}\,\left(-1\right)\ =\ \pi]


If you want it in degrees you can multiply by *[tex \Large \frac{180}{\pi}]


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