Thread: How should I solve this Trigonometric equation?

For math homework I have to solve the following equation. I know what the answer is, I just don't quite understand how to get there. Can you please help me?
So the equation is:

sin(2x - ⅓π) = -cos(x + ⅓π)

This is how I've solved it so far:

sin(2x - ⅓π) = -cos(x + ⅓π)
sin(2x - ⅓π) = cos(x + 4/3π)
sin(2x - ⅓π) = sin(x + 11/6π)

then using sin(A) = sin(B) gives A = B + k2π and A = π - B + K2π, I got:

2x - ⅓π = x + 11/6π + K2π and 2x - ⅓π = π - (x+11/6π) + k2π
x = 13/6π + k2π and 3x - ⅓π = -½π + k2π
x = 13/6π + k2π and x = - 1/6π + k ^. ⅔π

This I all understand. Now, however, we have to solve this on an interval of [0, 2π]

The answer: x = 1/6π and x = ½π and x = 7/6π and x = 11/6π

I don't understand how they got there. Could anyone please explain this to me in as much detail as possible?
Thank you so much!

You have chosen, which to me is, a strange method of solution, but the results look OK.
Simply substitute any integer values for k that will give you a value for x in the required range.
That's k=-1 for your first solution and k=1,2,3 for the second.