# Solving for X

• Apr 1st 2010, 05:00 AM
Punch
Solving for X
Given the curve $y=1-2cos^2x+cosx$ for $0. Calculate the values of x for which the curve meets the x-axis.

I don't know where I went wrong but this is my attempt, textbook answer is $x=\frac{4\pi}{3}$ and $\frac{2\pi}{3}$

Attempt

Meet x-axis, $y=0$

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

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

$cosx=\frac{-1}{2}[Q_2,Q_3] , cosx=1[Q_1,Q_4]$

$\alpha=2.0943rad$ & $\alpha=0$

$x=\pi-2.0943, x=\pi+2.0943, x=0(rej), x=2\pi(rej)$

= $1.05rad = 5.23rad$
• Apr 1st 2010, 05:52 AM
Quote:

Originally Posted by Punch
Given the curve $y=1-2cos^2x+cosx$ for $0. Calculate the values of x for which the curve meets the x-axis.

I don't know where I went wrong but this is my attempt, textbook answer is $x=\frac{4\pi}{3}$ and $\frac{2\pi}{3}$

Attempt

Meet x-axis, $y=0$

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

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

$cosx=\frac{-1}{2}[Q_2,Q_3] , cosx=1[Q_1,Q_4]$ your work is fine to here

$\alpha=2.0943rad$ & $\alpha=0$ these are 2 of 4 solutions which include 0 and 360 degrees.

$x=\pi-2.0943, x=\pi+2.0943, x=0(rej), x=2\pi(rej)$

= $1.05rad = 5.23rad$

Hi Punch,

Cosx is the horizontal co-ordinate of a point on the circle of radius =1 if you like, centred at the origin.

When you obtain arccos using the value -0.5,
your calculator will return one of two angles this corresponds to.

You wrote your angles values "without" $\pi$.

It's much simpler to approach this type of question,
by considering cosx to be a horizontal co-ordinate of value -0.5,
to the left of the origin.

This co-ordinate references 2 angles, one less than 180 degrees
and the other greater than 180 degrees.

If you calculate arccos(0.5) you obtain the acute angle that you add to or subtract from 180 degrees.

Note that the question excludes x=0 or x=360 degrees,
so it is asking you to find the 2 angles x, referenced by cosx=-0.5.
• Apr 1st 2010, 06:04 AM
Punch
so does that mean that my answer is right?
• Apr 1st 2010, 06:58 AM
Punch
thanks i understood it
• Apr 1st 2010, 01:03 PM
Hi Punch,

the first value of $\alpha$ that you calculated is one value of the angle x.
You should not be combining that value with $\pi$
• Apr 1st 2010, 06:40 PM
Punch
I was working out the value of the basic angle.
• Apr 2nd 2010, 02:22 AM
to get the "acute" angle, evaluate arccos(0.5), ie $cos^{-1}0.5$