# Math Help - Solve for X:

1. ## Solve for X:

Solve for X:

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

2. 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

3. ## I screwed Up

I screwed UP!
it should be

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

help is apreciated!

4. Use the same substitution. $u=\cos(x)$ You'll get a quadratic.