document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161423>Question 1161423</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve exactly, where possible, giving answers in radians. Approximate answers must be rounded to 2 decimal places: \r<BR>\n" );
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document.write( "sec^2 a+5 = 3tan^2 a </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #784959 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=784959\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the equation is sec^2(a) + 5 = 3*tan^2(a).<BR>\n" );
document.write( "secant is equal to 1/cos and tangent is equal to sin/cos, therefore:<BR>\n" );
document.write( "the equation becomes 1/cos^2(a) + 5 = 3*sin^2(a)/cos^2(a).<BR>\n" );
document.write( "multiply both sides of this equation by cos^2(a) to get:<BR>\n" );
document.write( "1 + 5*cos^2(a) = 3*sin^2(a)<BR>\n" );
document.write( "sin^2(a) is equal to 1 = cos^2(a).<BR>\n" );
document.write( "the equation becomes:<BR>\n" );
document.write( "1 + 5*cos^2(a) = 3*(1 - cos^2(a))<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( "1 + 5*cos^2(a) = 3 - 3*cos^2(a)<BR>\n" );
document.write( "subtract 1 from both sides of the equation and add 3*cos^2(a) to both sides of the equation and combine like terms to get:<BR>\n" );
document.write( "8*cos^2(a) = 2<BR>\n" );
document.write( "solve for cos^2(a) to get:<BR>\n" );
document.write( "cos^2(a) = 2/8 = 1/4<BR>\n" );
document.write( "solve for cos(a) to get:<BR>\n" );
document.write( "cos(a) = plus or minus 1/2.<BR>\n" );
document.write( "solve for a to get:<BR>\n" );
document.write( "a = plus or minus arccos(1/2)<BR>\n" );
document.write( "in degrees, a would be equal to 60 degrees for arccos(1/2) or 120 degrees for -arccos(1/2).<BR>\n" );
document.write( "60 degrees * pi / 180 = pi/3 in radians.<BR>\n" );
document.write( "120 degrees * pi / 180 = 2pi/3 in radians.<BR>\n" );
document.write( "those would be your answers after they're confirmed.<BR>\n" );
document.write( "to confirm, replace a with those answers after your set your calculator to radians and evaluate the original equation to see if it's true.<BR>\n" );
document.write( "the original equation is:<BR>\n" );
document.write( "sec^(a) + 5 = 3 * tan^2(a)]<BR>\n" );
document.write( "when a = pi/3 radians, that becomes:<BR>\n" );
document.write( "sec^2(pi/3) + 5 = 3 * tan^2(pi/3)<BR>\n" );
document.write( "if  your calculator doesn't do secant directly, than substitute 1/cos^2(pi/3) for sec^2(pi/3).<BR>\n" );
document.write( "you will get:<BR>\n" );
document.write( "9 = 9, confirming pi/3 is one of the angles.<BR>\n" );
document.write( "when a = 2pi/3 radians, that (the original equation) becomes:<BR>\n" );
document.write( "1/cos^2(2pi/3) + 5 = 3 * tan^2(2pi/3)<BR>\n" );
document.write( "you will get:<BR>\n" );
document.write( "9 = 9, confirming 2pi/3 is also one of the angles.<BR>\n" );
document.write( "note that a can also be 360 - 60 = 300 degrees * pi/180 = 5pi/3 and a can also be 180 + 60 = 240 * pi / 180 = 4pi/3.<BR>\n" );
document.write( "you can graph the original equations to see if they're equal at those angles.<BR>\n" );
document.write( "the graph confirms that sec^2(a) + 5 is equal to 3 * tan^2(a) for a = pi/3, 2pi/3, 4pi/3, 5pi/3 in the interval of a = 0 to 2pi.<BR>\n" );
document.write( "the graph looks like this:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2020/062001.jpg\" >\r<BR>\n" );
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document.write( "since you are not restricted to the interval from a = 0 to a = 2pi, then there are an infinite number of angles that will satisfy the equation.<BR>\n" );
document.write( "those angles would be:<BR>\n" );
document.write( "pi/3 plus or minus k * 2pi<BR>\n" );
document.write( "2pi/3 plus or minus k * 2pi<BR>\n" );
document.write( "4pi/3 plus or minus k * 2pi<BR>\n" );
document.write( "5pi/3 plus or minus k * 2pi<BR>\n" );
document.write( "all of those angles satisfy the equation.<BR>\n" );
document.write( "k is an integer greater than or equal to 0.<BR>\n" );
document.write( "as one example, take a = 5pi/3 and let k = plus or minus 5<BR>\n" );
document.write( "when k = 0, the angle is 5pi/3 which we already know satisfies the equation because 1/cos^2(5pi/3) + 5 = 9 and 3 * tan^2(5pi/3) = 9<BR>\n" );
document.write( "when k = -5, 5pi/3 becomes 5pi/3 - 5 * 2pi which becomes 5pi/3 - 10pi which becomes 5pi/3 - 30pi/3 which becomes -25pi/3.<BR>\n" );
document.write( "1/cos^2(-25pi/3) + 5= 9<BR>\n" );
document.write( "3 * tan^2(-25pi/3) = 9<BR>\n" );
document.write( "that angle satisfies the equation.<BR>\n" );
document.write( "when k = 5, 5pi/3 becomes 5pi/3 + 5 * 2pi = 5pi/3 + 10pi which becomes 5pi/3 + 30pi/3 which becomes 35pi/3.<BR>\n" );
document.write( "1/cos^(35pi/3) + 5 = 9<BR>\n" );
document.write( "3 * tan^2(35pi/3) = 9<BR>\n" );
document.write( "that angle also satisfies the equation.<BR>\n" );
document.write( "on a graph, that looks like this:\r<BR>\n" );
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document.write( " </SPAN>\n" );
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