# Thread: f(x) = 2 cos x

1. ## f(x) = 2 cos x

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)?

2. Can I see your attempts?

Originally Posted by Chop Suey
Can I see your attempts?
Our teacher often places a "challenge question" on the board for extra credit but does not take the time to show the class or give a hint.

Can you help me or not?

4. this is hardly a "challenge" question ... it should be clear that

$\cos{x} = -\frac{\sqrt{3}}{2}$

do you know your unit circle well enough to say what two values x satisfy the equation?

5. ## skeeter...

Originally Posted by skeeter
this is hardly a "challenge" question ... it should be clear that

$\cos{x} = -\frac{\sqrt{3}}{2}$

do you know your unit circle well enough to say what two values x satisfy the equation?
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.

6. ## my attempts..

Originally Posted by Chop Suey
Can I see your attempts?
My attempts are not even close to what must be done here.

7. Originally Posted by magentarita
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)?
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.

$f(x) = 2 \cos x$

(a) $f(x) = - \sqrt{3}$ means

$2 \cos x = - \sqrt{3}$

$\Rightarrow \cos x = - \frac {\sqrt{3}}2$

Now we recall from tables that we have (or should have) memorized before that $\cos \frac {\pi}6 = \frac {\sqrt{3}}2$

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

the reference angles for $\frac {\pi}6$ in the 2nd and 3rd quadrants are $\frac {5 \pi}6$ and $\frac {7 \pi}6$ respectively.

so your solutions are:

$x = \frac {5 \pi}6 + 2n \pi$ or $x = \frac {7 \pi}6 + 2k \pi$, for some integers $n$ and $k$

(b) Now we want $2 \cos x < - \sqrt{3}$

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

$\cos x < - \frac {\sqrt{3}}2$

similar to the above: $\implies x < \frac {5 \pi}6$ or $x < \frac {7 \pi}6$

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 $\frac {5 \pi}6$ into the original and see if the result is less than $-\sqrt{3}$, then plug in a number between $\frac {5 \pi}6$ and $\frac {7 \pi}6$, and check again. then a number to the right of $\frac {7 \pi}6$, and check again.

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

$\frac {5 \pi}6 < x < \frac {7 \pi}6$

8. ## Fabulous...

Originally Posted by Jhevon
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.

$f(x) = 2 \cos x$

(a) $f(x) = - \sqrt{3}$ means

$2 \cos x = - \sqrt{3}$

$\Rightarrow \cos x = - \frac {\sqrt{3}}2$

Now we recall from tables that we have (or should have) memorized before that $\cos \frac {\pi}6 = \frac {\sqrt{3}}2$

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

the reference angles for $\frac {\pi}6$ in the 2nd and 3rd quadrants are $\frac {5 \pi}6$ and $\frac {7 \pi}6$ respectively.

so your solutions are:

$x = \frac {5 \pi}6 + 2n \pi$ or $x = \frac {7 \pi}6 + 2k \pi$, for some integers $n$ and $k$

(b) Now we want $2 \cos x < - \sqrt{3}$

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

$\cos x < - \frac {\sqrt{3}}2$

similar to the above: $\implies x < \frac {5 \pi}6$ or $x < \frac {7 \pi}6$

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 $\frac {5 \pi}6$ into the original and see if the result is less than $-\sqrt{3}$, then plug in a number between $\frac {5 \pi}6$ and $\frac {7 \pi}6$, and check again. then a number to the right of $\frac {7 \pi}6$, and check again.

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

$\frac {5 \pi}6 < x < \frac {7 \pi}6$
I now can go into the textbook and search for similar questions and work them out following your math notes here.

Thanks.