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)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \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 \n" ); document.write( " \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. \n" ); document.write( " |