document.write( "Question 609385: How do you solve sin(x/2)-cosx=0 with a restriction of [0,2pi)? We are supposed to be using the half angle formula... \n" ); document.write( "
Algebra.Com's Answer #384129 by lwsshak3(11628)\"\" \"About 
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How do you solve sin(x/2)-cosx=0 with a restriction of [0,2pi)? We are supposed to be using the half angle formula
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\n" ); document.write( "cos half-angle formula:
\n" ); document.write( "cos x/2=√[(1+cosx)/2]
\n" ); document.write( "√[(1+cosx)/2]-sinx=0
\n" ); document.write( "√[(1+cosx)/2]=sinx
\n" ); document.write( "square both sides
\n" ); document.write( "(1+cosx)/2=sin^2x=1-cos^2x
\n" ); document.write( "1+cosx=2-2cos^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=π/3 and 5π/3
\n" ); document.write( "or
\n" ); document.write( "cosx+1=0
\n" ); document.write( "cosx=-1
\n" ); document.write( "x=π \n" ); document.write( "
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