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jdoe0001 Group TitleBest 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"
 9 months ago

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

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

surjithayer Group TitleBest 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=?
 9 months ago
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