Question

Trig help?

Answer 1

What do they mean solve the equation for 3x?

Answer 2

They mean solve for the `angle`.
The angle is the thing inside of the cosine function, ya?

Answer 3

\[\large\rm \cos(\color{orangered}{3x})=\cos(\color{orangered}{\theta})\]The entire thing inside of the cosine function is our angle.

Answer 4

\[\large\rm \cos(\theta)=-1\qquad\to\qquad \theta=?\]

Answer 5

pi?

Answer 6

Mmm good.\[\large\rm \cos(\color{orangered}{\theta})=-1\qquad\qquad\to\qquad\qquad \color{orangered}{\theta}=\pi\]Or it terms of our problem,\[\large\rm \cos(\color{orangered}{3x})=-1\qquad\qquad\to\qquad\qquad \color{orangered}{3x}=\pi\]That's what they want for part 1.

Answer 7

Thank you! And for Part II, x = pi/3 right? Is that the only solution?

Answer 8

Well part 2 we'll have to be a little careful.
3x=pi is the only solution of 3x in [0,2pi], yes.
But let's still look at the general solutions.\[\large\rm \cos(3x)=-1\qquad\qquad\to\qquad\qquad 3x=\pi+2k \pi\]

Answer 9

Dividing by 3,\[\large\rm x=\frac{\pi}{3}+\frac{2k \pi}{3}\]

Answer 10

So while we had only a single solution for 3x,
it looks like we'll probably have more for x, ya?

Answer 11

For part 1, would they want me to write the 2kpi on the end?
I'm a little confused on what the solutions will be... How would I figure that out?

Answer 12

For part 1,
you can look at this:\[\large\rm 3x=\pi+2k \pi\]and determine which k values will give you solutions.
k=0 gives us \(\large\rm 3x=\pi\)
k=1 gives us \(\large\rm 3x=3\pi\) but oops! This is not in [0,2pi].
So we've found our only solution for part 1: \(\large\rm 3x=\pi\)
No need to mention the general form in your answer.

Answer 13

For part 2,
you already determined one of the solutions.
It's the one that corresponds to k=0, ya?\[\large\rm x=\frac{\pi}{3}+0\]How bout k=1?

Answer 14

What is k? Could it be a decimal or a fraction?

Answer 15

Sorry :) k is an integer (positive or negative whole number).

Answer 16

k=1 is not between 0 and 2pi, so it wouldn't be part of the solution set right? for part II

Answer 17

What does the K stand for? How many full rotations it does?

Answer 18

k=1 doesn't give a solution between 0 and 2pi?
You sure about that?\[\large\rm x=\frac{\pi}{3}+\frac{2(1)\pi}{3}\]

Answer 19

Before we divided by 3, yes that's what it stood for.

Answer 20

Now that we've divided by 3, it stands for something different.
Looks like 1/3 rotations or something, ya?

Answer 21

My mind... exploding... okay, I think I got it...
hmmmm. So pi/3 and pi are both solutions... also 5pi/3? And that's it?

Answer 22

\[\large\rm x=\frac{\pi}{3}+\frac{2(2)\pi}{3}=\frac{5\pi}{3}\]Mmmm yes.. good good good.
The next one is like... 7pi/3 which is outside of what we want.

Answer 23

The fact that we `divided by 3` and ended up with 3 times as many solutions... I don't think that's a coincidence. That's probably how it will always work out. Don't quote me on that though :P

Answer 24

For part 1, would I put 3x = pi or 3x = pi + 2kpi ?

Answer 25

3x=pi

Answer 26

Why is that? :/

Answer 27

The same reason we didn't include any k's in solution to part 2, right? :)
We only care about the specific k's that give us solutions.

Answer 28

The only one that mattered for part 1 was k=0.

Answer 29

For part 2, k=0,1,2 gave us solutions.

Answer 30

Ahh I see...
thanks zepp :) You sure are Ze-best!

Answer 31

np Mabble :O