SOLUTION: 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.

Algebra ->  Trigonometry-basics -> SOLUTION: 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.      Log On


   

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.
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
cos%282x%29%2Bcos%28x%29-1=0
2cos%5E2%28x%29-1%2Bcos%28x%29-1=0
2cos%5E2%28x%29%2Bcos%28x%29-2=0

cos%28x%29=%28-1%2B-+sqrt%281-4%2A2%2A%28-2%29%29%29%2F4

cos%28x%29=%28-1%2B-+sqrt%289%29%29%2F4

cos%28x%29=%28-1%2B-+3%29%2F4

cos%28x%29=-1 or cos%28x%29=1%2F2
Notice where x points to on the unit circle for these cosines.
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.
... next value for x would be 7pi/3.

...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.