1. ## Solving for X

Given the curve $\displaystyle y=1-2cos^2x+cosx$ for $\displaystyle 0<x<2\pi$. 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 $\displaystyle x=\frac{4\pi}{3}$ and $\displaystyle \frac{2\pi}{3}$

Attempt

Meet x-axis,$\displaystyle y=0$

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

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

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

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

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

=$\displaystyle 1.05rad = 5.23rad$

2. Originally Posted by Punch
Given the curve $\displaystyle y=1-2cos^2x+cosx$ for $\displaystyle 0<x<2\pi$. 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 $\displaystyle x=\frac{4\pi}{3}$ and $\displaystyle \frac{2\pi}{3}$

Attempt

Meet x-axis,$\displaystyle y=0$

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

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

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

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

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

=$\displaystyle 1.05rad = 5.23rad$
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" $\displaystyle \pi$.

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.

3. so does that mean that my answer is right?

4. thanks i understood it

5. Hi Punch,

the first value of $\displaystyle \alpha$ that you calculated is one value of the angle x.
You should not be combining that value with $\displaystyle \pi$

6. I was working out the value of the basic angle.

7. Hi Punch,

to get the "acute" angle, evaluate arccos(0.5), ie $\displaystyle cos^{-1}0.5$

rather than arccos(-0.5).

Then you can add to and subtract from 180 degrees to obtain x.

When you evaluate arccos(-0.5), calculators typically give one of the
two angles x. You still need to calculate the other one.

8. Thanks, I spotted that