Question 267298
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\sin{\Theta}\ -\ 2\cos^2{\Theta}\ =\ -3]


We know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2{\phi}\ +\ \cos^2{\phi}\ =\ 1]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2{\phi}\ =\ 1\ -\ \sin^2{\phi}]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\sin{\Theta}\ -\ 2\left(1\ -\ \sin^2{\Theta}\right)\ =\ -3]


Let *[tex \Large x\ =\ \sin{\Theta}]


Then:


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


Collect terms/Standard form:


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


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


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


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


So


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin{\Theta}\ =\ -1]



So on the interval *[tex \LARGE 0\ \leq\ \Theta\ \leq\ 2\pi]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Theta\ =\ \frac{7\pi}{6}]


or 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Theta\ =\ \frac{11\pi}{6}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Theta\ =\ \frac{3\pi}{2}]


Check the unit circle:


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


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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