Hi folks,

Normally when solving trig equations, for example: $tan \ x = -\frac{1}{3}$, then $tan^{-1} (\frac{1}{3}) = 18.4$ then because the tangent is negative, solutions are found in the second and fourth quadrants. So given a range $0 \le x \le 360 $ the solutions are 161.6 degrees and 341.6 degrees.

Now, I am trying to solve a trig equation using the $t = tan(\frac{x}{2})$ substitution. In this:

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

So, I am trying to solve:

$7 \ cos \ x + 6 \ sin \ x = 2$

and using the substitution, I get $t = -\frac{1}{3}$ and $t = \frac{5}{3}$.

therefore

$\frac{x}{2} = - 18.4$ degrees $\Rightarrow x = - 36.9 $ degrees

As mentioned above, the solutions are in quad 2 and 4 so for the given range we should get x = 143.1 and x = 323.1

My maths books only gives x = 323.1 as a solution. For some reason, the solution in the second quad is ignored. Why would this be?

For $tan \frac{x}{2} = \frac{5}{3}$ we get x = 118.1 and this is again the only solution provided. I would expect a solution in quad 4 (298.1).

I am wondering whether the fact that $\frac{x}{2} = 59$ degrees is in quad 1 and x = 118.1 is in quad 2 is the reason.

Can anyone explain this?