document.write( "Question 871008: Find the solutions in the interval [0, 2pi)\r
\n" ); document.write( "\n" ); document.write( "2 cos 3x = 1\r
\n" ); document.write( "\n" ); document.write( "This is due tomorrow and I am very confused, any and all help is greatly appreciated :)
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Algebra.Com's Answer #525258 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
\"2cos%283x%29=1\"
\n" ); document.write( "\"cos%283x%29=1%2F2\"
\n" ); document.write( "The cosine function takes all the positive values between 0 and 1, without repeating in quadrant I and again in quadrant IV.
\n" ); document.write( "In quadrants II and III cosine has negative values.
\n" ); document.write( "In quadrant I (between 0 and \"pi%2F2\" , or if you prefer between\"0%5Eo\" and \"90%5Eo\" ),
\n" ); document.write( "we have \"cos%28pi%2F3%29=1%2F2\" .
\n" ); document.write( "In quadrant IV , \"-pi%2F3\" has a cosine of \"1%2F2\" too.
\n" ); document.write( "Since cosine has a period of \"2pi\" , adding \"2pi\" you get an angle with the same cosine.
\n" ); document.write( "So all the angles with a cosine of \"1%2F2\" can be expressed as
\n" ); document.write( "\"2k%2Api+%2B-+pi%2F3\"=\"%286k+%2B-+1%29pi%2F3\" , for any integer \"k\"
\n" ); document.write( "If \"cos%283x%29=1%2F2\" , then \"3x=%286k+%2B-+1%29pi%2F3\" --> \"x=%286k+%2B-+1%29pi%2F9\" .
\n" ); document.write( "\"%286%2A0+-+1%29pi%2F9=-pi%2F9%3C0\" and \"%286%2A3%2B1%29pi%2F9=highlight%2819pi%2F9%3E2pi%29\" are outside the interval \"%22%5B+0+%2C%22\"\"2pi\"\"%22%29%22\" .
\n" ); document.write( "The solutions in the interval \"%22%5B+0+%2C%22\"\"2pi\"\"%22%29%22\" are
\n" ); document.write( "\"%286%2A0+%2B+1%29pi%2F9=highlight%28pi%2F9%29\" , \"%286%2A1+-+1%29pi%2F9=highlight%285pi%2F9%29\" , \"%286%2A1+%2B+1%29pi%2F9=highlight%287pi%2F9%29\" , \"%286%2A2+-+1%29pi%2F9=highlight%2811pi%2F9%29\" , \"%286%2A2%2B1%29pi%2F9=highlight%2813pi%2F9%29\" , \"%286%2A3+-+1%29pi%2F9=highlight%2817pi%2F9%29\" .
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