document.write( "Question 630941: how do you solve sin(x/2)+cosx=0 with a restriction of [0,2pi)? \n" ); document.write( "
Algebra.Com's Answer #397249 by lwsshak3(11628)\"\" \"About 
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how do you solve sin(x/2)+cosx=0 with a restriction of [0,2pi)?
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\n" ); document.write( "Identity: sin(x/2)=±√[(1-cosx)/2]
\n" ); document.write( "sin(x/2)+cosx=0
\n" ); document.write( "sin(x/2)=cosx
\n" ); document.write( "±√[(1-cosx)/2]=cosx
\n" ); document.write( "square both sides
\n" ); document.write( "(1-cosx)/2=cos^2x
\n" ); document.write( "1-cosx=2cos^2x
\n" ); document.write( "2cos^2x+cosx-1=0
\n" ); document.write( "(2cosx+1)(cosx-1)=0
\n" ); document.write( "..
\n" ); document.write( "2cosx+1=0
\n" ); document.write( "cosx=-1/2
\n" ); document.write( "x=2π/3 and 4π/3 (in quadrants II and III where cos<0)
\n" ); document.write( "or
\n" ); document.write( "cosx-1=0
\n" ); document.write( "cosx=1
\n" ); document.write( "x=0\r
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