Suppose that f(x) = 2 cos x.

(a) Solve for f(x) = -sqrt{3}

(b) For what values of x is f(x) < -sqrt{3} on the interval

[0, 2pi)?

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- Aug 31st 2008, 06:25 AM #1

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- Aug 31st 2008, 06:44 AM #2

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- Aug 31st 2008, 01:00 PM #3

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- Aug 31st 2008, 01:08 PM #4

- Aug 31st 2008, 05:13 PM #5

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## skeeter...

skeeter,

I thank you for taking time out to reply but honestly if my questions are too easy and boring, then what's stopping you from skipping my questions? I am trying to pass my pre-calculus class and trying hard to grasp math concepts but your replies of late are discouraging.

- Aug 31st 2008, 05:26 PM #6

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- Aug 31st 2008, 06:01 PM #7
you need to start showing your attempts. otherwise, we won't know what to help you with and have no idea if you are improving or not. i think the comments made in this thread so far are fair. the questions the users ask, though they may seem offensive, are to help you. please be cooperative. nevertheless, here's what you want.

(a) means

Now we recall from tables that we have (or should have) memorized before that

Now, if we have a negative answer, namely, , it means we are in the 2nd or 3rd quadrant. (this is why skeeter mentioned the unit circle)

the reference angles for in the 2nd and 3rd quadrants are and respectively.

so your solutions are:

or , for some integers and

(b) Now we want

solving in the same way we would if we had an equal sign, we have

similar to the above: or

now we draw a number line and plot both those points. then we test the regions to see if we have the inequality.

that is, plug in a number to the left of into the original and see if the result is less than , then plug in a number between and , and check again. then a number to the right of , and check again.

you will realize only the middle number works, so that your solution is:

- Sep 1st 2008, 03:14 AM #8

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