Question 346794
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Recall that *[tex \Large \cos^2\varphi\ +\ \sin^2\varphi\ =\ 1], hence:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \pm\sqrt{1\ -\ \frac{7}{16}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \pm\sqrt{\frac{9}{16}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \pm\frac{3}{4}]


But you restricted *[tex \Large x] the the third quadrant, *[tex \Large \pi\ <\ x\ <\ \frac{3\pi}{2}].  Sin is negative in the third quadrant, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ -\frac{3}{4}]


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