# Math Help - Trigo

1. ## Trigo

2. Originally Posted by ose90
The function f is defined by f(x)=cos x + (1/2) cos 2x , 0 <= x <= 2pi

find all values of x in the form of k pi, with k correct to one decimal places when f(x)=0

My solution:

cosx + 1/2 cos 2x = 0
2 cos x + cos 2x = 0
2 cos x + (2cos^2 x-1) = 0
2 cos ^2 x + 2 cos x - 1 = 0

Using quadratic formula, i get x = -1/2 +- (sqrt3)/2

How should I express them in term of pi?
You only need to solve $\cos x = \frac{\sqrt{3} - 1}{2}$ (why?).

Note that you only need them in the form k pi, with k correct to one decimal place. So use technology to get decimal approximations of all solutions and then solve for k, correct to one decimal place.

Eg. One solution is x = 1.19606. So $1.19606 = k \pi => k = \, .....$, correct to one decimal place.

3. Originally Posted by ose90
Deleting your question and proclaiming that you have the answer is NOT good form.

1. Other members might be interested in the question and its solution. MHF would be a pretty crummy resource if everyone did what you've done.

2. The answer you've got might be wrong. Surely getting some sort of 'peer review' of your answer is desirable.

3. Some members might have valuable contributions to make (like alternative methods of solution, interesting extensions to the original question etc.).

There are other reasons but these are the three most important, in my opinion anyway.

4. Originally Posted by mr fantastic
Deleting your question and proclaiming that you have the answer is NOT good form.

1. Other members might be interested in the question and its solution. MHF would be a pretty crummy resource if everyone did what you've done.

2. The answer you've got might be wrong. Surely getting some sort of 'peer review' of your answer is desirable.

3. Some members might have valuable contributions to make (like alternative methods of solution, interesting extensions to the original question etc.).

There are other reasons but these are the three most important, in my opinion anyway.