QUOTE(Betinho @ Apr 27 2006, 09:47 PM)

Hmm.. I think I understand. But what about this one?

**cosx - root3sinx = 1**

k = 2

alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5

-> x - 300 = 60, 300?

I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Say your looking for values of x that satisfy this equation with the range:

Now we have:

Now if we took the normal values for the inverse cosine of 0.5 we would get the angle of 60

^{o}, and we would say that within the range the values are 60

^{o} and 300

^{o}However, we are adding 300

^{o} onto the angle, and if we add 300

^{o} onto 300

^{o} then we get a value outwith our range.

However, adding 300

^{o} onto 60

^{o} gives us a value still within the range so it's fine to use this

There is another value for x though, and if you think of this as coming from the cosine graph

*before* zero (because it repeats, just as there is a repeat of the graph after 360 there is one before zero). So if we took -60 (i.e. the equivalent of the quadrant four angle in the graph before zero) as one of our values for x, added 300 onto this and got 240 this would give us a value still within our range. Also, if we took the quadrant one angle from our graph before zero, which is -300 (i.e. -360 + 60), and add 300 onto this we get zero which is still in our range.

So from this we get the results:

x = 0, 240, 360

That's how I do it anyway, not sure about how good a method it is