document.write( "Question 970967: i got 2n(pi) +or- pi/3 and 2n(pi) +or- pi as solutions for the equation cos x+cos 2x=0. but the book's answer says (2n+1)pi/3. I would like to know how could that be. \n" ); document.write( "
Algebra.Com's Answer #593597 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
\"cos%282x%29%2Bcos%28x%29-1=0\"
\n" ); document.write( "\"2cos%5E2%28x%29-1%2Bcos%28x%29-1=0\"
\n" ); document.write( "\"2cos%5E2%28x%29%2Bcos%28x%29-2=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"cos%28x%29=%28-1%2B-+sqrt%281-4%2A2%2A%28-2%29%29%29%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "\"cos%28x%29=%28-1%2B-+sqrt%289%29%29%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "\"cos%28x%29=%28-1%2B-+3%29%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "\"cos%28x%29=-1\" or \"cos%28x%29=1%2F2\"
\n" ); document.write( "Notice where x points to on the unit circle for these cosines.
\n" ); document.write( "pi/3, 3pi/3, 5pi/3, and then as you continue to rotate in increments of to maintain those, you see difference of 2pi/3 each time.
\n" ); document.write( "... next value for x would be 7pi/3.\r
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\n" ); document.write( "\n" ); document.write( "...as you keep on thinking on this, notice that x will be ODD increments of pi/3. Now observe that your numerator for x IS AN ODD NUMBER.
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