Solve Mcos^2(x) + Ncos(x) – 3 = 0 for M & N | Trig Help

[Mathhelp77]

Homework Statement

If cos(x) = -¾ or cos = ½ then the value of M and N in the equation Mcos^2(x) + Ncos(x) – 3 = 0 are what?

Homework Equations

Identities I can use:
csc(x) = 1/sin(x)
cot(x) = 1/tan(x)
sec(x) = 1/cos(x)
tan(x) = sin(x)/cos(x)
cot(x) = cos(x)/sin(x)
sin^2(x) +cos^2(x) = 1
1 +tan^2(x) = sec^2(x)
1 +cot^2(x) = csc^2(x)
sin(A+B) = (sinA)(cosB) + (cosA)(sinB)
sin(A-B) = (sinA)(cosB) - (cosA)(sinB)
cos(A+B) = (cosA)(cosB) - (sinA)(sinB)
cos(A-B) = (cosA)(cosB) + (sinA)(sinB)
sin(2A) = 2(sinA)(cosA)
cos(2A) = cos^2A - sin^2A

The Attempt at a Solution

 

[lewando]

cos = ½

What is that? Welcome to PF. You need to attempt a solution before we can really help.

 

[Mathhelp77]

Alright :) I have kind of a few different things I've been working on and I'm not sure which is right or if any of them are...

I considered using the pythagorean identity to get
-Msin^2(x) + Ncos(x) - 2 = 0

and that is my best guess but that doesn't really make sense cause then I am bringing sin which might make it more difficult :S So... perhaps I have to see what it could factor into?

 

[eumyang]

This is not really a trig question, but more of an algebra question. I assume you know how to solve quadratics by factoring. Here's an example:

$$\begin{aligned} 3y^2 - y - 4 &= 0 \\ (3y - 4)(y + 1) &= 0 \\ 3y - 4 &= 0 \rightarrow y = 4/3 \\ y + 1 &= 0 \rightarrow y = -1 \end{aligned}$$

But suppose I gave you the solutions y = 4/3 and y = -1 and I want you to find the quadratic with those solutions. We go "backwards":
$$\begin{aligned} y = 4/3 \rightarrow 3y = 4 \rightarrow 3y - 4 &= 0 \\ y = -1 \rightarrow y + 1 &= 0 \\ (3y - 4)(y + 1) &= 0 \\ 3y^2 - y - 4 &= 0 \end{aligned}$$

You have to do something similar here. You're given the "solutions":
cos x = -3/4 and cos x = 1/2.
Pretend that "cos x" is the variable. Go "backwards" and find the quadratic in terms of cos x. Equate the coefficients with
$$M\cos^2 x + N\cos x - 3 = 0$$
to find M and N. That's it! No identities needed.

 

[Mathhelp77]

okay that makes sense! Thanks a lot you were a huge help!