Question 1132562: solve: 1+cos(6x)-2cos(4x)=0 on [0 ,pi]
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
It's not a formal algebraic solution, however....
Note that since the expression is to be equal to 0, it seems the values of cos(6x) and cos(4x) have to be rational numbers. Assuming that, there are two ways to make the expression equal to 0:
(1) cos(6x) = 1; cos(4x) = 1 --> 1+1-2(1) = 0
(2) cos(6x) = 0; cos(4x) = 1/2 --> 1+0-2(1/2) = 0
For answers of the first type, the angles 6x and 4x must both be multiples of 2pi. On the given closed interval [0,pi], there are two values of x for which that is true: 0 and pi.
First set of answers: {0,pi}
For answers of the second type, 4x must be pi/3 or 5pi/3, or either of those plus a multiple of 2pi. On the given closed interval, there are four values of x for which that is true: pi/12, 5pi/12, 7pi/12, and 11pi/12. For each of those values of x, cos(6x)=0, as required, so these are all solutions to the equation.
Second set of answers: {pi/12, 5pi/12, 7pi/12, 11pi/12}
The solutions found by this method are then the following:
{0, pi/12, 5pi/12, 7pi/12, 11pi/12, pi}
Graphing the equation with a graphing calculator confirms that those are all the solutions in the given interval.
Answer by ikleyn(52781)  (Show Source):
|
|
|
