I hope I don't sound silly (again :p) however I'm very confused by this question.

Solve algebraically:

2 cos (x - 13°) = , x € [0°, 720°]

I'm not even sure why there is a euro sign in there... (Doh)

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- Jul 16th 2010, 10:05 AMwiseguyTrigonometric equation
I hope I don't sound silly (again :p) however I'm very confused by this question.

Solve algebraically:

2 cos (x - 13°) = , x € [0°, 720°]

I'm not even sure why there is a euro sign in there... (Doh) - Jul 16th 2010, 10:11 AMAckbeet
That should be $\displaystyle x\in[0^{\circ},720^{\circ}],$ not the Euro sign. A typo, I'm sure. The set membership symbol is nearly universal.

Question: what's on the RHS of your equation there? I see this:

$\displaystyle 2\cos(1-13^{\circ})=$

That's not an equation, or even a well-defined formula of any sort. - Jul 16th 2010, 10:14 AMwiseguy
x is between 0 and 720 degrees?

How should I go about answering this question? - Jul 16th 2010, 10:15 AMAckbeet
Correct. x is in the interval between 0 and 720 degrees, inclusive.

I wouldn't even start answering this question until I had a RHS to the equation. - Jul 16th 2010, 11:07 AMwiseguy
Here's what I got so far:

cos(x-13°)=√3/2

x-13°=arccos√3/2

x-13°=3.14/6

x=13°+3.14/6

??? - Jul 16th 2010, 11:12 AMAckbeet
That's certainly a start. I would probably not work in mixed "units": use all degrees or all radians. So, you've essentially found one solution (you should also use $\displaystyle \pi$ instead of an approximation). But there might be more. How would you go about finding them?

- Jul 16th 2010, 11:23 AMwiseguy
Is it safe to convert the 13 degrees to radians?

- Jul 16th 2010, 11:32 AMAckbeet
Of course you can. Now, mind you, the original problem has degrees. Most of the time, you want to give answers in a unit system consistent with what's given in the problem statement. So I would convert the pi/6 into degrees, to be consistent with the problem statement.

- Jul 16th 2010, 11:46 AMwiseguy
cos(x-13°)=√3/2

x-13°=arccos√3/2

x-13°=3.14/6

x=13°+30°

x=43°

plus pi a few times, and the answer is 43, 223, 403, and 583 degrees... how does it look - Jul 16th 2010, 11:48 AMAckbeet
Well, check your answers. Do they work in a calculator? (Make sure it's in degrees mode!)

- Jul 16th 2010, 11:50 AMAckbeet
I predict the 403 will work, but not the 223 or the 583.

- Jul 16th 2010, 11:50 AMwiseguy
Ok, this is the type of problem that needs to be checked for extraneous solutions?

- Jul 16th 2010, 11:53 AMAckbeet
All problems should be double-checked to make sure the answer makes sense. That's part of problem-solving in general. It's especially important when you have to square an equation (you might introduce extraneous roots).

So, what do you get? I would agree that there are four solutions, and I think you've found exactly two of them. - Jul 16th 2010, 11:56 AMwiseguy
since x is limited to [0°, 720°], wouldn't the fifth solution (763°) be kicked out?

- Jul 16th 2010, 11:58 AMAckbeet
You are correct not to continue past 720 degrees. However, I don't agree with your solutions 223 or 583 degrees. Plug them into the equation and see if they satisfy it!