# Trigonometry Algebraic problem

1. Jul 11, 2013

### hunter45

Hello,

Could you please help me with a problem that I saw in my exam? I have tried to solve it but end up getting one solution which is not correct. There happens to be 2 solutions to the problem which appear periodically.

The problem has to be algebraically solved without the use of a graphing calculator so I would appreciate if working is given.

Equation 1: p(x) = 5000 cos [∏/2 (x-1)] + 6000
Equation 2: p(x) = 15000 cos [∏/2 (x+((∏/2)-1))] + 25000

Last edited: Jul 11, 2013
2. Jul 11, 2013

### Millennial

What does the problem ask for? The value of x? And are the functions in eqn 1 and 2 different? You denoted one of them by p and the other one by P.

3. Jul 11, 2013

### hunter45

Yes, solve for x. All I know is that you must equate them. I do not know where to go from there.

4. Jul 11, 2013

### Millennial

Equate them and do the necessary simplifications, then use the cosine sum formulae to expand the cosines.

5. Jul 11, 2013

### hunter45

Thats what I did but the answer is still incorrect. Could you please work it out so I can see how to do it correct.

6. Jul 11, 2013

### SteamKing

Staff Emeritus
No, at PF, it is the policy for you to show your work. We will examine it and note any flaws in your reasoning and/or calculations.

7. Jul 12, 2013

### volta14641

I don't think that this is a trigonometric equation so to speak in the sense of solving for x, what it looks like to me is a Cosine graph in which the vertical shift is +6000, the amplitude is 5000 and the period for cosine is "2pi over b". "b" in this case is "Pi over 2." "2Pi" divided by "Pi over 2" is 4. So that is your period for this cosine graph. This problem requires you to make a table which is almost impossible for me to describe in words but basically the top row would be 0, pi/2, pi, 3pi/2, and 2pi. I'm sorry this reads confusing because I don't know where the Pi button is yet. In trigonometry pi over 2 also refers to 90 degrees and pi refers to 180 degrees.