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jdoe0001
 11 months ago
Best ResponseYou've already chosen the best response.0\(\bf sin (2x) =\cfrac{1}{2}\quad \textit{take }sin^{1} \\ \quad \\ sin^{1}[sin ({\color{red}{ 2x}})]=sin^{1}\left(\frac{1}{2}\right)\implies {\color{red}{ 2x}}=sin^{1}\left(\frac{1}{2}\right)\) solve for "x"

kb6
 11 months ago
Best ResponseYou've already chosen the best response.0It says to solve the equation on the interval 0<x<2pi

jdoe0001
 11 months ago
Best ResponseYou've already chosen the best response.0yes so... what did you get for "x"?

surjithayer
 11 months ago
Best ResponseYou've already chosen the best response.0\[\sin 2x=\frac{ 1 }{ 2 }=\sin \frac{ \pi }{6 }=\sin \left( \pi+\frac{ \pi }{6 } \right),\sin \left( 2\pi\frac{ \pi }{ 6 } \right),\sin \left( 2\pi+\pi+\frac{ \pi }{ 6 } \right),\] \[\sin \left( 2\pi+2\pi\frac{ \pi }{6 } \right)\] 2x=? x=?
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