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...
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...
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.
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?
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.
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.