document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=27169>Question 27169</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please can you help me to find the general solution in radians of the following equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin+2x++-+1+=+cos+2x ALIGN=MIDDLE  ALT=\"sin+2x++-+1+=+cos+2x\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I think I need to use a trig identity but I'm not sure which one or how?\r<BR>\n" );
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document.write( "Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #14781 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=14781\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sin 2x  - 1 = cos 2x<BR>\n" );
document.write( "sin 2x - cos 2x = 1...divide throughout with square root of sum of squares of coefficients of sin(2x) and cos(2x) ...we get sq.rt(1^1+1^2)=sq.rt(2)..dividing with sq.rt(2)<BR>\n" );
document.write( "if we put sin(1/sqrt(2))=pi/4 = cos(pi/4)..we get<BR>\n" );
document.write( "cos(pi/4)sin(2x)+sin(pi/4)cos(2x)=1<BR>\n" );
document.write( "sin(pi/4+2x)=1=sin(pi/2)<BR>\n" );
document.write( "general solution is <BR>\n" );
document.write( "(pi/4)+2x=n*(pi)+{(-1)^n}*(pi/2)<BR>\n" );
document.write( "x=(1/2){n*(pi)-(pi/4)+((-1)^n)*(pi/2)}\r<BR>\n" );
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