Question 1178292
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Let *[tex \Large u\ =\ \sin(2\theta)] then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 4u\ -\ 1\ =\ 0]


Solve for *[tex \LARGE u_1] and *[tex \LARGE u_2] with the quadratic formula,


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(2\theta_1)\ =\ u_1]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(2\theta_2)\ =\ u_2]



So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta_1\ =\ \frac{\sin^{-1}(u_1)}{2}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta_2\ =\ \frac{\sin^{-1}(u_2)}{2}]


You can do your own arithmetic and calculator work.

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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