Thread: 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.

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

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

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!

Originally Posted by classicstrings

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

if x is acute or obtuse (case i)

or otherwise (case ii)

Case i
So



Case ii

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

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!