document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1032167>Question 1032167</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve 2cos^2 - 5cos(x) +2 =0 for principal values of x. Express the solution(s) in degrees. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #646813 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=646813\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Solve 2cos^2 - 5cos(x) + 2 = 0 for principal values of x. Express the solution(s) in degrees.<BR>\n" );
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document.write( "2cos^2 -4cos -cos + 2 = 0<BR>\n" );
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document.write( "2cos(cos-2)-(cos-1) = 0<BR>\n" );
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document.write( "(cos-2)(2cos-1) = 0<BR>\n" );
document.write( "cos(x) = 2 or cos(x) = 1/2<BR>\n" );
document.write( "Acceptable solution::<BR>\n" );
document.write( "cos(x) = 1/2 implies x = 60 degrees<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( "  </SPAN>\n" );
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