1. ## Inverse cosine

cos2x = 3/4

So to find x, I would go: (cos-1 (3/4)) / 2

This answer gives x = 20.7

But the memo gives x = 0.36 and 2.78.

How did they get to those 2 values?

2. ## Re: Inverse cosine

Also, there are infinite number of possible such x:

$\displaystyle 2x \in \left\{\pm \cos^{-1}(3/4) + 2 \pi n \ | \ n \in \mathbb{Z} \right\}$, so

$\displaystyle x \in \left\{\pm \frac{\cos^{-1}(3/4)}{2} + \pi n \ | \ n \in \mathbb{Z} \right\}$.

The books 2nd answer corresponds to $\displaystyle \pi - \frac{\cos^{-1}(3/4)}{2}$.

In a sense, there are just two "different" answers (given by that $\displaystyle \pm$), because all the rest are just integer mulitples of $\displaystyle \pi$ added to those two.

3. ## Re: Inverse cosine

Ah, yes I forgot that the sum includes the limits 0 <= x <= pi.

So how would I go about working it out?

4. ## Re: Inverse cosine

Ah, yes I forgot that the sum includes the limits 0 <= x <= pi.
Which sum?

5. ## Re: Inverse cosine

Did you understand what I wrote?

Do you know where "$\displaystyle 2x \in \left\{\pm \cos^{-1}(3/4) + 2 \pi n \ | \ n \in \mathbb{Z} \right\}$" comes from?

It's algebraically how you get all solutions. (In this case, we had to divide by 2 everywhere from that initial set.) Then to see how many solutions fall within some interval stated by your problem, you simply examine your solution set and algebraically decide which ones are in the desired range. That typically involves "getting an inequality for n" and solving it.

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Here that would be $\displaystyle 0 \le \cos^{-1}(3/4) + \pi n \le \pi$ and $\displaystyle 0 \le -\cos^{-1}(3/4) + \pi m \le \pi$.

Since $\displaystyle \cos^{-1}$ is in $\displaystyle [0, \pi]$, for the first case must have $\displaystyle n=0$ only, since if you added $\displaystyle \pi$ to something positive you'd jump ahead of $\displaystyle \pi$, and likewise if you subtracted a $\displaystyle \pi$, since you're not bigger than $\displaystyle \pi$, you'd end up negative.

For simlar reasons, $\displaystyle m = 1$.

So you should hopefully just "see it". However, in algebraic detail, solving those inequalities you get:

$\displaystyle -\frac{\cos^{-1}(3/4)}{\pi} \le n \le \frac{\pi - \cos^{-1}(3/4)}{\pi}$ and $\displaystyle \frac{\cos^{-1}(3/4)}{\pi} \le m \le \frac{\pi + \cos^{-1}(3/4)}{\pi}$

Now you could put that into your calculator to get: $\displaystyle -.23 \le n \le .77$ and $\displaystyle .23 \le m \le 1.23$, so n = 0 and m = 1.

Or you could note that, since $\displaystyle \cos^{-1}(3/4)$ is in $\displaystyle (0, \pi)$, it follows that $\displaystyle 0 < \frac{\cos^{-1}(3/4)}{\pi} < 1$.

Thus

$\displaystyle -1<-\frac{\cos^{-1}(3/4)}{\pi} \le n \le \frac{\pi - \cos^{-1}(3/4)}{\pi} < 1$ and $\displaystyle 0< \frac{\cos^{-1}(3/4)}{\pi} \le m \le \frac{\pi + \cos^{-1}(3/4)}{\pi}< 1+1 = 2$,

so again n = 0 and m = 1.
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I'll give a visual description - maybe it'll make sense, maybe not:
The full solution is an infinite set of points. Imagine them highlighted on a number line. The solution in a range will just be a "window" on the number line. However, that "2x" means that the final x you get will "bring more solutions into your window", because to go from 2x to x is to multiply by 1/2, which "shrinks the number line".

6. ## Re: Inverse cosine

Ok, I understood what you said there. So, my answer of 20.7 is just one of those valid numbers on the line, but it doesn't fall within the interval. So how do I get those answers of x = 0.36 and 2.78?

7. ## Re: Inverse cosine

Originally Posted by yorkey
Ah, yes I forgot that the sum includes the limits 0 <= x <= pi.
So how would I go about working it out?

If the range is really $\displaystyle 0\le x\le \pi$ .
The other answer is $\displaystyle \pi -\frac{\arccos\left(\frac{3}{4}\right)}{2}$.

8. ## Re: Inverse cosine

I'm afraid I still don't understand. Here's the question:

9. ## Re: Inverse cosine

Originally Posted by yorkey
I'm afraid I still don't understand. Here's the question:
Let this be a lesson to you. Always always post the entire question.
Having seen the whole question, I agree with the first reply. There are two answers in the range $\displaystyle 0\le x\le \pi$.