1. ## Trigonometric equation

I hope I don't sound silly (again ) 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...

2. 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.

3. x is between 0 and 720 degrees?

4. 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.

5. 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

???

6. 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?

7. Is it safe to convert the 13 degrees to radians?

8. 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.

9. 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

10. Well, check your answers. Do they work in a calculator? (Make sure it's in degrees mode!)

11. I predict the 403 will work, but not the 223 or the 583.

12. Ok, this is the type of problem that needs to be checked for extraneous solutions?

13. 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.

14. since x is limited to [0°, 720°], wouldn't the fifth solution (763°) be kicked out?

15. 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!

Page 1 of 2 12 Last