# Solve for X:

• November 7th 2006, 04:50 PM
Mr_Green
Solve for X:
Solve for X:

3cos(x) + 3 = -2cos^2(x), for x E [-pi, 3pi]
• November 7th 2006, 04:53 PM
topsquark
Quote:

Originally Posted by Mr_Green
Solve for X:

3cos(x) + 3 = -2cos^2(x), for x E [-pi, 3pi]

$2cos^2(x) + 3cos(x) + 3 = 0$

It might be easier for you to define y = cos(x), then look at:

$2y^2 + 3y + 3 = 0$

You can use the quadratic equation to solve this, discard any solutions that are greater than 1 or less than -1, then take the inverse cosine.

The problem is that this quadratic has no real solutions. Thus you have no real solutions for x. Is there a typo perhaps?

-Dan
• November 7th 2006, 07:27 PM
Mr_Green
I screwed Up
I screwed UP!
it should be

3cos(x) + 3 = 2 - 2cos^2(x), for x E [-pi, 3pi]

help is apreciated!
• November 7th 2006, 08:17 PM
Jameson
Use the same substitution. $u=\cos(x)$ You'll get a quadratic.