Question 633779
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2(x)\ +\ 3\sin(x)\ =\ 0]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\varphi)\ +\ \sin^2(\varphi)\ =\ 1]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(1\ -\ \sin^2(x)\right)\ +\ 3\sin(x)\ =\ 0]


Let *[tex \LARGE u\ =\ sin(x)], substitute, and arrange the quadratic into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(1\ -\ u^2\right)\ +\ 3u\ =\ 0]


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


Solve the factorable quadratic:


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


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


Substitute back, i.e. *[tex \LARGE \sin(x)\ =\ u].  But discard *[tex \LARGE u\ =\ \sin(x)\ =\ 2] because 2 is not in the domain of the sine function.


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ -\frac{1}{2}]


Refer to the unit circle, recalling that the sine values are the *[tex \LARGE y] coordinates of the intersection of the terminal ray with the unit circle.


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


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