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Atkinsoha
PreCalc & Trig, picture below, please show your work.. Really frustrated with this homework. :/
use sinx=2sin(x/2)cos(x/2) then factorize
Maybe I should tell you this: I haven't been in class for over a week. I don't know any of this stuff.. so be as simple and thorough as you can, thanks.
\[\cos(\frac{ x }{ 2 })-sinx=0\] use the identity sin2x=2sinxcosx \[\cos(x/2)-2\sin(x/2)\cos(x/2)=0\] \[\cos(x/2)(1-\sin(x/2))=0\] cos(x/2)=0 or 1-sin(x/2)=0 \[\cos(x/2)=0 => x/2=\pi/2 \] \[x=\pi\] \[1-\sin(\frac{ x }{ 2 })=0\]
\[\sin(\frac{ x }{ 2 })=1 => \frac{ x }{ 2 }=\pi/2\] \[x=\pi\]
the solution is \[\pi\]