Given the curve for . 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 and

Attempt

Meet x-axis,

&

=

Printable View

- Apr 1st 2010, 06:00 AMPunchSolving for X
Given the curve for . 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 and

__Attempt__

Meet x-axis,

&

= - Apr 1st 2010, 06:52 AMArchie Meade
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" .

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, 07:04 AMPunch
so does that mean that my answer is right?

- Apr 1st 2010, 07:58 AMPunch
thanks i understood it

- Apr 1st 2010, 02:03 PMArchie Meade
Hi Punch,

the first value of that you calculated is one value of the angle x.

You should not be combining that value with - Apr 1st 2010, 07:40 PMPunch
I was working out the value of the basic angle.

- Apr 2nd 2010, 03:22 AMArchie Meade
Hi Punch,

to get the "acute" angle, evaluate arccos(0.5), ie

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. - Apr 2nd 2010, 07:45 AMPunch
Thanks, I spotted that :)