document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197640>Question 1197640</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve for θ. 0° ≤ θ ≤ 360°.<BR>\n" );
document.write( "1. (sin^2)θ + (1/(sin^2)θ) + sinθ + (1/sinθ) = 4<BR>\n" );
document.write( "I tried substituting m for sinθ but after simplifying I get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3%2Bm%2B1-4m%5E2=0 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3%2Bm%2B1-4m%5E2=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >. I'm not sure what to do next? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #831004 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3835)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=831004\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%5E2%28theta%29+%2B+1%2F%28sin%5E2%28theta%29%29+%2B+sin%28theta%29+%2B+1%2F%28sin%28theta%29%29+=+4 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E2+%2B+1%2F%28m%5E2%29+%2B+m+%2B+1%2Fm+=+4 ALIGN=MIDDLE  ALT=\"m%5E2+%2B+1%2F%28m%5E2%29+%2B+m+%2B+1%2Fm+=+4\"   DISABLED_style= \"max-width:90%; height:auto;\" > Let m = sin(theta)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4+%2B+1+%2B+m%5E3+%2B+m+=+4m%5E2 ALIGN=MIDDLE  ALT=\"m%5E4+%2B+1+%2B+m%5E3+%2B+m+=+4m%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > Multiply each term by the LCD m^2 to clear out the fractions.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3%2Bm%2B1-4m%5E2+=+0 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3%2Bm%2B1-4m%5E2+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+0 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "It appears you followed these steps, or similar to them, since you arrived at what I got shown above. <BR>\n" );
document.write( "The steps are there to help any future student down the road with a similar problem. \r<BR>\n" );
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document.write( "The task is now solving <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+0 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > for m.\r<BR>\n" );
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document.write( "Use the rational root theorem to find the only possible rational roots are: -1 and 1.\r<BR>\n" );
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document.write( "Test each possible root. If you plugged in m = -1, then you should find <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" > turns into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%29%5E4%2B%28-1%29%5E3-4%28-1%29%5E2%2B%28-1%29%2B1+=+-4 ALIGN=MIDDLE  ALT=\"%28-1%29%5E4%2B%28-1%29%5E3-4%28-1%29%5E2%2B%28-1%29%2B1+=+-4\"   DISABLED_style= \"max-width:90%; height:auto;\" >. This shows m = -1 is NOT a root of the fourth degree polynomial. A root is some input that leads to an output of zero.\r<BR>\n" );
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document.write( "Try m = 1 and you should get 0 as a result.<BR>\n" );
document.write( "This shows m = 1 is a root of the fourth degree polynomial.<BR>\n" );
document.write( "This then means (m-1) is a factor of m^4+m^3-4m^2+m+1\r<BR>\n" );
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document.write( "Use either polynomial long division or synthetic division to find that<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+%28m-1%29%28m%5E3%2B2m%5E2-2m-1%29 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+%28m-1%29%28m%5E3%2B2m%5E2-2m-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I skipped showing the steps, but let me know if you need to see them. \r<BR>\n" );
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document.write( "Then use the rational root theorem on <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E3%2B2m%5E2-2m-1 ALIGN=MIDDLE  ALT=\"m%5E3%2B2m%5E2-2m-1\"   DISABLED_style= \"max-width:90%; height:auto;\" > to find that m = 1 is a root of that as well.\r<BR>\n" );
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document.write( "It turns out m = 1 is a double root of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Use polynomial long division or synthetic division to find that<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E3%2B2m%5E2-2m-1+=+%28m-1%29%28m%5E2%2B3m%2B1%29 ALIGN=MIDDLE  ALT=\"m%5E3%2B2m%5E2-2m-1+=+%28m-1%29%28m%5E2%2B3m%2B1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Overall, we can say <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+%28m-1%29%28m%5E3%2B2m%5E2-2m-1%29 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+%28m-1%29%28m%5E3%2B2m%5E2-2m-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+%28m-1%29%28m-1%29%28m%5E2%2B3m%2B1%29 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1+=+%28m-1%29%28m-1%29%28m%5E2%2B3m%2B1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Each m-1 factor leads to m = 1 as a root<BR>\n" );
document.write( "Which in turn means <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28theta%29+=+1 ALIGN=MIDDLE  ALT=\"sin%28theta%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Use a chart, calculator, or the unit circle to find the only solution  to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28theta%29+=+1 ALIGN=MIDDLE  ALT=\"sin%28theta%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" > is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=theta+=+matrix%281%2C2%2C90%2Cdegrees%29 ALIGN=MIDDLE  ALT=\"theta+=+matrix%281%2C2%2C90%2Cdegrees%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > where <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+%3C=+theta+%3C=+360 ALIGN=MIDDLE  ALT=\"0+%3C=+theta+%3C=+360\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "You'll need the quadratic formula to compute the roots of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E2%2B3m%2B1=0 ALIGN=MIDDLE  ALT=\"m%5E2%2B3m%2B1=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I'll skip the steps, but you should have these two roots:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m+=+%28-3%2Bsqrt%285%29%29%2F2 ALIGN=MIDDLE  ALT=\"m+=+%28-3%2Bsqrt%285%29%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m+=+%28-3-sqrt%285%29%29%2F2 ALIGN=MIDDLE  ALT=\"m+=+%28-3-sqrt%285%29%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "They approximate to <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2Bsqrt%285%29%29%2F2+=+-0.381966 ALIGN=MIDDLE  ALT=\"%28-3%2Bsqrt%285%29%29%2F2+=+-0.381966\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3-sqrt%285%29%29%2F2+=+-2.618034 ALIGN=MIDDLE  ALT=\"%28-3-sqrt%285%29%29%2F2+=+-2.618034\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Therefore, we have these additional equations to solve for theta<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28theta%29+=+-0.381966 ALIGN=MIDDLE  ALT=\"sin%28theta%29+=+-0.381966\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28theta%29+=+-2.618034 ALIGN=MIDDLE  ALT=\"sin%28theta%29+=+-2.618034\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "But wait, the second equation isn't possible since the range of sine is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%3C=+sin%28theta%29%3C=1 ALIGN=MIDDLE  ALT=\"-1%3C=+sin%28theta%29%3C=1\"   DISABLED_style= \"max-width:90%; height:auto;\" >. <BR>\n" );
document.write( "The lowest sine can get is -1, which means we can't reach -2.618034 unless we involve complex numbers. <BR>\n" );
document.write( "I'll assume your teacher is working with the real number set only.<BR>\n" );
document.write( "In short, we can rule out <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28theta%29+=+-2.618034 ALIGN=MIDDLE  ALT=\"sin%28theta%29+=+-2.618034\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "I'll leave the task of solving <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28theta%29+=+-0.381966 ALIGN=MIDDLE  ALT=\"sin%28theta%29+=+-0.381966\"   DISABLED_style= \"max-width:90%; height:auto;\" > to the student. <BR>\n" );
document.write( "Hint: there are 2 solutions in the interval <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+%3C=+theta+%3C=+360 ALIGN=MIDDLE  ALT=\"0+%3C=+theta+%3C=+360\"   DISABLED_style= \"max-width:90%; height:auto;\" >; one in quadrant III and the other in quadrant IV<BR>\n" );
document.write( "I recommend using the unit circle.\r<BR>\n" );
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document.write( "I also recommend graphing software like Desmos or GeoGebra to confirm that you found the correct roots of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1 ALIGN=MIDDLE  ALT=\"m%5E4%2Bm%5E3-4m%5E2%2Bm%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" > (there are 3 distinct roots). Furthermore, that graphing software should also verify each solution for the trig functions in terms of theta.<BR>\n" );
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