document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=609369>Question 609369</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #384131 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=384131\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula.\r<BR>\n" );
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document.write( "cos half-angle formula:<BR>\n" );
document.write( "cos x/2=&#8730;[(1+cosx)/2]<BR>\n" );
document.write( "&#8730;[(1+cosx)/2]-sinx=0<BR>\n" );
document.write( "&#8730;[(1+cosx)/2]=sinx<BR>\n" );
document.write( "square both sides<BR>\n" );
document.write( "(1+cosx)/2=sin^2x=1-cos^2x<BR>\n" );
document.write( "1+cosx=2-2cos^2x<BR>\n" );
document.write( "2cos^2x+cosx-1=0<BR>\n" );
document.write( "(2cosx-1)(cosx+1)=0<BR>\n" );
document.write( "..<BR>\n" );
document.write( "2cosx-1=0<BR>\n" );
document.write( "cosx=1/2<BR>\n" );
document.write( "x=&#960;/3 and 5&#960;/3<BR>\n" );
document.write( "or<BR>\n" );
document.write( "cosx+1=0<BR>\n" );
document.write( "cosx=-1<BR>\n" );
document.write( "x=&#960; </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );