Solve for X:

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

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- Nov 7th 2006, 04:50 PMMr_GreenSolve for X:
Solve for X:

3cos(x) + 3 = -2cos^2(x), for x E [-pi, 3pi] - Nov 7th 2006, 04:53 PMtopsquark
$\displaystyle 2cos^2(x) + 3cos(x) + 3 = 0$

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

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

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 - Nov 7th 2006, 07:27 PMMr_GreenI screwed Up
I screwed UP!

it should be

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

help is apreciated! - Nov 7th 2006, 08:17 PMJameson
Use the same substitution. $\displaystyle u=\cos(x)$ You'll get a quadratic.