document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=484085>Question 484085</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve 2cos x+2cos 2x=0 on the interval [0,2pi) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #331326 by </B><A HREF=/tutors/aboutme.mpl?userid=Joker><B>Joker(6)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Joker><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=331326\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First use the double angle formula to get 2cos x+2(2cos^2 x-1)=0 then distribute and simplify by dividing by 2 to get 2cos^2 x+cos x-1=0 then factor into (2cos x-1)(cos x+1)=0 and solve for x using the unit circle where cos x = 1/2 and -1 which is at pi/3,5pi/3 and pi. </SPAN>\n" );
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