Question 333055
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Trigonometry.


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


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\sin^2(x)\ =\ \cos^2(x)]


Use the Pythagorean Identity *[tex \Large \left(\cos^2(\varphi)\ +\ \sin^2(\varphi)\ =\ 1\text{ hence }\cos^2(\varphi)\ =\ 1\ -\ \sin^2(\varphi)\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\sin^2(x)\ =\ 1\ -\ \sin^2(x)]


A little algebra


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\sin^2(x)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(x)\ =\ \frac{1}{5}]


Take the square root, considering only the positive root because
both *[tex \Large sin] and *[tex \Large cos] are positive on the interval *[tex \Large 0^\circ\ <\ x\ <\ 90^\circ].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \sqrt{\frac{1}{5}}]


Rationalize the denominator, just for tidiness sake:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \frac{sqrt{5}}{5}]


Then use the inverse function to solve for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \sin^{-1}\left(\frac{sqrt{5}}{5}\right)]


And finally a little calculator work, in degrees mode because of the nature of the interval given, and we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ \approx\ 26.6^\circ]


Check


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(26.6^\circ)\ \approx\ 0.448]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(26.6^\circ)\ \approx\ 0.894]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ \cdot\ 0.448\ =\ 0.896]


Close enough.


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