Question 888366
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2x\ =\ \cos^2x]


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


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin x\ =\ \pm\frac{\sqrt{2}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \arcsin\left(\frac{\sqrt{2}}{2}\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \arcsin\left(-\frac{\sqrt{2}}{2}\right)]


There are two places on the unit circle where the *[tex \Large y]-coordinate of the point of intersection with the terminal ray of the angle is equal to *[tex \Large \frac{\sqrt{2}}{2}], and two places where the *[tex \Large y]-coordinate of the point of intersection with the terminal ray of the angle is equal to *[tex \Large -\frac{\sqrt{2}}{2}]. These are the four elements of the solution set to your equation presuming that you are interested in the solution set over the interval *[tex \Large 0\ \leq\ x\ <\ 2\pi]


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