Hi folks,

I am trying to solve the following equation $9cos x - 8 sin x = 12$ using the "t" substitution i.e. $t = \tan \frac{x}{2}$ which gives

$ tan \ x = \dfrac{2t}{1 - t^2}, \ sin \ x = \dfrac{2t}{1 + t^2}, \ cos \ x = \dfrac{1 - t^2}{1 + t^2} $

the result is $t = -\dfrac{1}{3}$ and $t = -\dfrac{3}{7}$

In my last post skeeter explained how to get solutions in the range $0\circ \le x \le 360\circ$ and I thought I had this issue locked down, but no!

This question asks for solutions in the range $-360\circ \le x \le 360\circ$ and gives the following answers: $-46.4\circ, -36.9\circ, 313.6\circ \ and \ 323.1\circ$

This is confusing because -46.4 occupies exactly the same position as 313.6 and -36.9 is the same as 323.1. The solutions overlap: one can say -46.4 degrees or 313.6 degrees (both the same place in quad 4) so I would expect one or the other, not both.

Also, since $tan^{-1} (-\frac{1}{3})$ gives a solution in quad4, then there should also be a solution in quad 2 (i.e. 143.1) but this is not given. I had a similar problem recently and the explanation was that one of my two solutions was out of range. In this case I can't see that this is a problem. So I would be very glad if someone could help me out here!