document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1199491>Question 1199491</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Q:1- Solve Cos(x)= 6x?<BR>\n" );
document.write( "Q:2- Solve Cos(3x) = 2x? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #833410 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</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=833410\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll focus on question 1 only.\r<BR>\n" );
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document.write( "cos(x) = 6x<BR>\n" );
document.write( "cos(x)-6x = 0\r<BR>\n" );
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document.write( "Let f(x) = cos(x)-6x\r<BR>\n" );
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document.write( "The derivative is f ' (x) = -sin(x)-6\r<BR>\n" );
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document.write( "Let's say we were to guess at random that x = 0 is a root of f(x).\r<BR>\n" );
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document.write( "Clearly that isn't the case because <BR>\n" );
document.write( "f(x) = cos(x)-6x<BR>\n" );
document.write( "f(0) = cos(0)-6*0<BR>\n" );
document.write( "f(0) = 1-6*0<BR>\n" );
document.write( "f(0) = 1-0<BR>\n" );
document.write( "f(0) = 1<BR>\n" );
document.write( "If that x value was a root, then the output should be f(x) = 0.\r<BR>\n" );
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document.write( "We'll use Newton's Method<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx\">https://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx</a><BR>\n" );
document.write( "to help narrow down on a root based on this input guess.<BR>\n" );
document.write( "a = old value = 0<BR>\n" );
document.write( "b = new value<BR>\n" );
document.write( "b = a - f(a)/( f ' (a) )<BR>\n" );
document.write( "b = 0 - f(0)/( f ' (0) )<BR>\n" );
document.write( "b = 0 - 1/( -6 )<BR>\n" );
document.write( "b = 1/6\r<BR>\n" );
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document.write( "Then,<BR>\n" );
document.write( "f(x) = cos(x)-6x<BR>\n" );
document.write( "f(1/6) = cos(1/6)-6*1/6<BR>\n" );
document.write( "f(1/6) = -0.01386 approximately<BR>\n" );
document.write( "Your calculator needs to be in radian mode, since the derivative is dependent on radian mode.\r<BR>\n" );
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document.write( "The input x = 1/6 doesn't lead to f(x) = 0, but it appears we're getting closer.\r<BR>\n" );
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document.write( "Repeat another iteration of Newton's Method.<BR>\n" );
document.write( "Use the result of the previous section (1/6) as the input value this time.<BR>\n" );
document.write( "a = old value = 1/6<BR>\n" );
document.write( "b = new value<BR>\n" );
document.write( "b = a - f(a)/( f ' (a) )<BR>\n" );
document.write( "b = 1/6 - f(1/6)/( f ' (1/6) )<BR>\n" );
document.write( "b = 1/6 - (-0.01386)/( -6.16590 )<BR>\n" );
document.write( "b = 0.16442\r<BR>\n" );
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document.write( "Then repeat again<BR>\n" );
document.write( "a = old value = 0.16442<BR>\n" );
document.write( "b = new value<BR>\n" );
document.write( "b = a - f(a)/( f ' (a) )<BR>\n" );
document.write( "b = 0.16442 - f(0.16442)/( f ' (0.16442) )<BR>\n" );
document.write( "b = 0.16442 - (-0.00001)/( -6.16368 )<BR>\n" );
document.write( "b = 0.164418\r<BR>\n" );
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document.write( "Once more<BR>\n" );
document.write( "a = old value = 0.164418<BR>\n" );
document.write( "b = new value<BR>\n" );
document.write( "b = a - f(a)/( f ' (a) )<BR>\n" );
document.write( "b = 0.164418 - f(0.164418)/( f ' (0.164418) )<BR>\n" );
document.write( "b = 0.164418 - (0.00000578)/( -6.16367821 )<BR>\n" );
document.write( "b = 0.16441893775174<BR>\n" );
document.write( "b = 0.16442\r<BR>\n" );
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document.write( "Fairly quickly we converge on the approximate root x = 0.16442\r<BR>\n" );
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document.write( "Newton's method isn't perfect. One glaring flaw is that if we were to pick a guess that leads to a horizontal tangent, then we'll get a division by zero error. Another problem is that the guess may lead to diverging values that do not narrow in on a particular root. The good news is that neither issue occurs with this particular function.\r<BR>\n" );
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document.write( "If you aren't familiar with calculus and/or derivatives, then here is an alternative pathway to finding roots of a function.\r<BR>\n" );
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document.write( "Once again we start with the input guess x = 0<BR>\n" );
document.write( "It leads to f(x) = 1 which is positive.<BR>\n" );
document.write( "Your calculator needs to be in radian mode.\r<BR>\n" );
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document.write( "Then notice how x = 1 leads to f(x) = -5.459698 approximately. The value itself doesn't matter. <BR>\n" );
document.write( "The key takeaway is that this value is negative.\r<BR>\n" );
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document.write( "f(x) is positive when x = 0<BR>\n" );
document.write( "f(x) is negative when x = 1<BR>\n" );
document.write( "The f(x) curve is continuous meaning that f(x) = 0 must be somewhere in the middle of these x values. <BR>\n" );
document.write( "Refer to the intermediate value theorem.\r<BR>\n" );
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document.write( "A good educated guess is the midpoint<BR>\n" );
document.write( "(0+1)/2 = 1/2 = 0.5<BR>\n" );
document.write( "Then determine that f(0.5) = -2.122417 approximately\r<BR>\n" );
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document.write( "f(x) is positive when x = 0<BR>\n" );
document.write( "f(x) is negative when x = 0.5<BR>\n" );
document.write( "At least one root is between these x values<BR>\n" );
document.write( "The midpoint is (0+0.5)/2 = 0.25<BR>\n" );
document.write( "Then f(0.25) = -0.531088 which is negative.\r<BR>\n" );
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document.write( "f(x) is positive when x = 0<BR>\n" );
document.write( "f(x) is negative when x = 0.25<BR>\n" );
document.write( "At least one root is between these x values<BR>\n" );
document.write( "The midpoint is (0+0.25)/2 = 0.125<BR>\n" );
document.write( "Then f(0.125) = 0.242198 which is positive.\r<BR>\n" );
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document.write( "f(x) is positive when x = 0.125<BR>\n" );
document.write( "f(x) is negative when x = 0.25<BR>\n" );
document.write( "At least one root is between these x values<BR>\n" );
document.write( "The midpoint is (0.125+0.25)/2 = 0.1875<BR>\n" );
document.write( "Then f(0.1875) = -0.142527 which is negative.\r<BR>\n" );
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document.write( "f(x) is positive when x = 0.125<BR>\n" );
document.write( "f(x) is negative when x = 0.1875<BR>\n" );
document.write( "At least one root is between these x values<BR>\n" );
document.write( "The midpoint is (0.125+0.1875)/2 = 0.15625<BR>\n" );
document.write( "Then f(0.15625) = 0.050318 which is positive.\r<BR>\n" );
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document.write( "f(x) is positive when x = 0.15625<BR>\n" );
document.write( "f(x) is negative when x = 0.1875<BR>\n" );
document.write( "At least one root is between these x values<BR>\n" );
document.write( "The midpoint is (0.15625+0.1875)/2 = 0.171875<BR>\n" );
document.write( "Then f(0.171875) = -0.045984 which is negative.\r<BR>\n" );
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document.write( "f(x) is positive when x = 0.15625<BR>\n" );
document.write( "f(x) is negative when x = 0.171875<BR>\n" );
document.write( "At least one root is between these x values<BR>\n" );
document.write( "The midpoint is (0.15625+0.171875)/2 = 0.1640625<BR>\n" );
document.write( "Then f(0.1640625) = 0.002197 which is positive.\r<BR>\n" );
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document.write( "Keep this process going until you narrow in on the approximate root x = 0.16442 mentioned earlier.\r<BR>\n" );
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document.write( "This process is known as the Bisection Method. <BR>\n" );
document.write( "We're basically zooming in on a piece of the number line by cutting it in half each time, then focusing on the half that has the root. <BR>\n" );
document.write( "The process is repeated until you achieve the desired level of precision.\r<BR>\n" );
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document.write( "For further reading, check out a similar concept of the Babylonian Method for calculating square roots by hand.\r<BR>\n" );
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document.write( "As mentioned by the tutor @ikleyn, it is best to use quick numeric methods that the calculator/computer will handle. <BR>\n" );
document.write( "Spreadsheet software is another option.\r<BR>\n" );
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document.write( "Desmos and GeoGebra are two free graphing calculator apps that I recommend. WolframAlpha is another good choice.<BR>\n" );
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