Question 611643
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I'm going to presume you mean "for the principal values of the inverse cosine"


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


This is much easier to see if you use substitution.  Let *[tex \LARGE u\ =\ \cos(x)]


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


Rearrange terms:


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


Factor a *[tex \LARGE u] from the first two terms and a 1 from the second pair of terms:


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


Then factor out *[tex \LARGE 2u^2\ -\ 1]


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


Hence


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ \pm\frac{\sqrt{2}}{2}]


The principal values of *[tex \LARGE \cos^{-1}(z)] are in the closed interval *[tex \LARGE -\frac{\pi}{2}\ \leq\ z\ \leq\ \frac{\pi}{2}]


Since *[tex \LARGE \cos(x)\ \geq\ 0\ \forall\,x\,\in\ -\frac{\pi}{2}\ \leq\ z\ \leq\ \frac{\pi}{2}]


Discard both negative roots.  Use the unit circle to find the two values of *[tex \LARGE x] in the principal interval that satisfy the equation.  Hint:  cosine  is the *[tex \LARGE x] coordinate of a point on the unit circle.  Anothe hint: Angle values in the 4th quadrant from the point of view of the pricipal interval are negative values.


<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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