Question 606881
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(x)\ =\ -\cos(x)]


on


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0\ \leq\ x\ \leq\ 2\pi]


Put the quadratic into standard form:


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


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\left(\cos(x)\ +\ 1\right)\ =\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ = -1]


Use the unit circle to find the points where the *[tex \LARGE x]-coordinate is equal to either 0 or -1 on the specified interval (namely, once around the 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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