# Thread: Solving for X, and General solutions?

1. ## Solving for X, and General solutions?

I have 2cos^2(x)+cos(x)=1 and I have to solve for x within a standard circle(0-2pi).

Then I have to find the general solution for the above equation and 2tan(x) +SqRt(12)=0(which I previously solved for.

2. Originally Posted by Blahdkm
I have 2cos^2(x)+cos(x)=1 and I have to solve for x within a standard circle(0-2pi).

[snip]
Simply subtract one from both sides and the factor the quadratic

$\displaystyle 2cos^2x+cosx-1=0$

$\displaystyle (2cosx-1)(cosx+1)=0$

now see that cosx=1/2 or -1

what angles do these correspond to?

Originally Posted by Blahdkm
[snip]

Then I have to find the general solution for the above equation and 2tan(x) +SqRt(12)=0(which I previously solved for.
$\displaystyle 2tanx=-\sqrt{12}$
$\displaystyle 2tanx=-2\sqrt{3}$
$\displaystyle tanx=-\sqrt{3}$
i only know of two angles.......
any questions?

thanx mr. fantastic

3. $\displaystyle 2tanx=-\sqrt{12}$
$\displaystyle 2tanx=-2\sqrt{3}$
$\displaystyle tanx=-\sqrt{3}$
i only know of two angles.......
$\displaystyle x=-\frac{\pi}{3}+n\pi$
where n is an integer.

4. Thank you two SO MUCH for the help! Any chance of assistance finding the general solution of 2cos^2(x) + cos(x)=1?

5. Originally Posted by Blahdkm
Thank you two SO MUCH for the help! Any chance of assistance finding the general solution of 2cos^2(x) + cos(x)=1?
$\displaystyle cos x=\frac{1}{2} \implies x=2n\pi \pm \frac{\pi}{3}$
$\displaystyle cos x=-1 \implies x=2n\pi \pm \pi$