# Stuck on a few multi choice...

• Oct 22nd 2006, 12:08 AM
classicstrings
Stuck on a few multi choice...
The number of solutions to the equation x = 5cos2x where x is a subset of the R is?

If tan(x) = 1/a
sin^2(x) = 1/b
cos(x) = ?

I tried using the identities but I cant seem to get right answer.

The intensity, I, of a light bumb is I = k/(x^2) where x = distance from bulb. THe rate at which the intensity changes with respect to distance when x = 5 is?

THis one is dI/dx right? So I get -2k/5, which is wrong.
• Oct 22nd 2006, 12:33 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
The intensity, I, of a light bumb is I = k/(x^2) where x = distance from bulb. THe rate at which the intensity changes with respect to distance when x = 5 is?

THis one is dI/dx right? So I get -2k/5, which is wrong.

dI/dx = -2k/(x^3)

so dI/dx at x=5 is: -2k/5^3 = -2k/225.

RonL
• Oct 22nd 2006, 12:36 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
The number of solutions to the equation x = 5cos2x where x is a subset of the R is?

If tan(x) = 1/a
sin^2(x) = 1/b
cos(x) = ?

I tried using the identities but I cant seem to get right answer.

I'm not quite sure what the question is here. The number of solutions of:

x= 5 cos(2x) (or is that x = 5 cos^2 (x) ?)

is not related as far as I can see to your list of candidate solutions.

RonL
• Oct 22nd 2006, 12:46 AM
classicstrings
Quote:

Originally Posted by CaptainBlack
I'm not quite sure what the question is here. The number of solutions of:

x= 5 cos(2x) (or is that x = 5 cos^2 (x) ?)

is not related as far as I can see to your list of candidate solutions.

RonL

Oops sorry they are 2 different questions!

For the first the solns are 3,5,7,9 but i am not sure how to work out the answer.

Then the 2nd q, I can't do either. Thanks CB!
• Oct 22nd 2006, 01:08 AM
Glaysher
Quote:

Originally Posted by classicstrings

If tan(x) = 1/a
sin^2(x) = 1/b
cos(x) = ?

$\displaystyle \tan x =\frac{\sin x}{\cos x}$

$\displaystyle \sin x = \frac{1}{\sqrt{b}}$ if x is acute or obtuse (case i)

or $\displaystyle \sin x = -\frac{1}{\sqrt{b}}$ otherwise (case ii)

Case i
So $\displaystyle \frac{1}{a} = \frac {\frac{1}{\sqrt{b}}}{\cos x}$

$\displaystyle \cos x = \frac {a}{\sqrt{b}}$

Case ii
$\displaystyle \cos x = -\frac {a}{\sqrt{b}}$
• Oct 22nd 2006, 01:44 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
Oops sorry they are 2 different questions!

For the first the solns are 3,5,7,9 but i am not sure how to work out the answer.

Then the 2nd q, I can't do either. Thanks CB!

As 5 cos(2x) lies between +/-5, all the roots of x=5 cos(2x) lie between
+/-5. So if you graph the two function you can just count the roots.

See attachment (looks like 7 roots to me).

RonL
• Oct 22nd 2006, 05:25 AM
classicstrings
Quote:

Originally Posted by CaptainBlack
dI/dx = -2k/(x^3)

so dI/dx at x=5 is: -2k/5^3 = -2k/225.

RonL

Dang, my brain is fried, such a silly error. Thanks for the help Glaysher too!