Question

Given the Problem find all solutions to in the interval of (0, 2pi)

(3cos(x)-1)(-cos(x)-1)

I know that (-cos(x)-1) = pi + 2kpi
k being an element of Integers

For
(3cos(x)-1)

I know one answer is
cos^-1(1/3) + 2kpi
k being an element of Integers

How do I find the other solution to
(3cos(x)-1)

Bonus marks if you can explain to me how to find both solutions for
(3sin(x)-1)

Answer 1

Why can't anyone help me with this? I seriously have looked all over the interwebs and my textbook only shows the easiest problems for example

Answer 2

Ok, so you've got most of it. But remember, you are only looking at the interval from (0,2pi), so you've only found two solutions total, you don't really need the + 2(k)pi part.

Then, you know that your function is symmetric about x = pi. In other words, before, you did 0 + cox^(-1)(1/3), so now do 2pi - cox^(-1)(1/3) for your second answer.

Those are all three of your answers.

Answer 3

To be clear, the one answer I provided you with is the third and last answer you are looking for. (but this is obvious).

I have a different way of thinking about this problem that should make more sense, hopefully.

For (3sin(x)-1)=0, if you picture the normal 3sin(x) graph: you see it has zeros at 0, and pi, and 2pi, yes? So, if you are moving it down by 1, you take sin^(-1)(1/3), as before, you see that the zero at (0,0) has moved slightly to the right. The precise amount is sin^(-1)(1/3) so your answer is 0 + sin^(-1)(1/3). You see that your zero at (pi,0) has moved slightly to the left, again the precise amount is sin^(-1)(1/3), so your answer is pi - sin^(-1)(1/3). Finally, you see your zero at 2pi has moved to the right by sin^(-1)(1/3). Alas, 2pi + sin^(-1)(1/3) is greater than pi, so this is not an answer within your bounds.

Answer 4

Thanks :D