document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1090692>Question 1090692</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve/simplify this.  Thanks.\r<BR>\n" );
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document.write( "Solve:  \r<BR>\n" );
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document.write( "2sin^2(theta) - cos(theta) - 1 = 0, on [0,2pi]\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #705165 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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/cgi-bin/show-face.mpl?user=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=705165\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For simplicity, we will replace theta with x.<BR>\n" );
document.write( "Using the trig identity, sin^2(x) = 1- cos^2(x), we have<BR>\n" );
document.write( "2(1-cos^2(x)) - cos(x) + 1 = 0<BR>\n" );
document.write( "2cos^2(x) + cos(x) - 1 = 0<BR>\n" );
document.write( "which factors as:<BR>\n" );
document.write( "(2cos(x)-1)(cos(x)+1) = 0 <BR>\n" );
document.write( "Thus cos(x) = -1 and cos(x) = 1/2<BR>\n" );
document.write( "If cos(x) = -1 -> x = pi<BR>\n" );
document.write( "If cos(x) = -1/2 -> x = 2pi/3 and 4pi/3 on the interval [0,2pi] </SPAN>\n" );
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