Question

How can i go about solving this sin^2 (x/2)= 1/2, 0 <= X <= 2PI

Answer 1

\[\sin^2 (\frac{ x }{ 2})=\frac{ 1 }{ 2 }\]
\[\sin (\frac{ x }{ 2 })=+-\frac{ \sqrt{2} }{ 2 }\]
take the positive value first
\[\sin(\frac{ x }{ 2 }) = \frac{ \sqrt{2} }{ 2 }\]
---> \[\frac{ x }{ 2 }= \frac{ \pi }{ 4 } or \frac{ 7\pi }{ 4 }\]
if \[\frac{ x }{ 2 }= \frac{ \pi }{ 4 }---> x = \frac{ \pi }{ 2 } + 2k\]
do the same with other values, I mean with 7pi/4 and then take negative value of sin (x/2)