Question 1014769
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2\cos^2(A)\ =\ 1\ +\ 2\sin(A)\cos(A)]


I presume you want to solve for A.  In the future, be specific about what you want; none of us are mind readers.


Add *[tex \Large -1] to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2(A)\ -\ 1\ =\ 2\sin(A)\cos(A)]


But *[tex \Large 2\cos^2(A)\ -\ 1\ = \cos(2A)] and *[tex \Large 2\sin(A)\cos(A)\ =\ \sin(2A)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(2A)\ =\ \sin(2A)]


Square both sides:


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


Use the Pythagorean Identity:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2(2A)\ =\ 1]


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2A\ =\ \cos^{-1}\left(\pm\frac{\sqrt{2}}{2}\right)\ =\ \frac{\pi}{4}\ +\ \frac{k\pi}{2}\ \forall\ k\ \in\ \mathbb{Z}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{\pi}{8}\ +\ \frac{k\pi}{4}\ \forall\ k\ \in\ \mathbb{Z}]


But if *[tex \Large k] is odd, the root is extraneous, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{\pi}{8}\ +\ \frac{k\pi}{2}\ \forall\ k\ \in\ \mathbb{Z}]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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