Question 1117815
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan^2\(x)\ =\ 1\ +\ \sec(x)]


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(x)\ =\ \frac{\cos^3(x)\ +\ \cos^2(x)}{\cos(x)}]


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


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


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


Let *[tex \Large u\ =\ \cos(x)]


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


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 1]


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


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


Hence


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


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


Now all you need is the two angles in the desired interval that have a cosine value of negative one-half.  Then you have to decide whether the stated interval is inclusive of the low end, the high end, both, or neither.


In other words, does the phrase "between 0 and 360 degrees" mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0^\circ\ <\ x\ <\ 360^\circ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0^\circ\ \leq\ x\ <\ 360^\circ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0^\circ\ <\ x\ \leq\ 360^\circ]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0^\circ\ \leq\ x\ \leq\ 360^\circ]


Because in the first case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^{-1}(1)\ \ ] Does not exist


In the second case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^{-1}(1)\ =\ 0^\circ]


In the third case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^{-1}(1)\ =\ 360^\circ]


And in the fourth case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^{-1}(1)\ =\ 0^\circ\ \ ] or *[tex \LARGE \cos^{-1}(1)\ =\ 360^\circ]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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