Question 596831
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First of all, I fixed your equation.  *[tex \LARGE \cos(x^2)] means the cosine of the angle *[tex \LARGE x^2], whereas  *[tex \LARGE \cos^2(x)] is the squared value of the cosine function at *[tex \LARGE x].  Very different things and they didn't give you


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos(x^2)\ -\ 5\cos(x)\ +\ 2\ =\ 0]


in a Precalculus book. 


The easiest way to see what is going on in this equation is to use a simple substitution.


Let *[tex \LARGE u\ =\ \cos(x)] and substitute.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2u^2\ -\ 5u\ +\ 2\ =\ 0]


Factor the quadratic in *[tex \LARGE u]


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


Apply the Zero Product Rule:


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


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 2]


Now recall the RANGE of the cosine function, i.e. *[tex \LARGE -1\ \leq\ \cos(\varphi)\ \leq\ 1].  That means that *[tex \LARGE u\ =\ 2] is an extraneous root and can be discarded.


Hence, substituting back


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


The principal values of the argument for cosine are in the interval


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0\ \leq\ \cos(x)\ \leq\ \pi]


Find the two angles in the unit circle where the *[tex \LARGE x]-coordinate is equal to *[tex \LARGE \frac{1}{2}] and select the one that is in the principal value interval.


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


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


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