# Thread: further issue with the t substitution

1. ## further issue with the t substitution

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!

2. ## Re: further issue with the t substitution

$-360 < x < 360 \implies -180 < \dfrac{x}{2} < 180$

$\tan\left(\dfrac{x}{2}\right) = -\dfrac{1}{3} \implies 90 < \dfrac{x}{2} <180$ or $-90 < \dfrac{x}{2} < 0$

For x/2 in quad II ...

$\dfrac{x}{2} = 180 + \arctan\left(-\dfrac{1}{3}\right) \implies x = 360 + 2\arctan\left(-\dfrac{1}{3}\right) = 323.1$

For x/2 in quad IV ...

$\dfrac{x}{2} = \arctan\left(-\dfrac{1}{3}\right) \implies x = 2\arctan\left(-\dfrac{1}{3}\right) = -36.9$

Now do the same for $\tan\left(\dfrac{x}{2}\right) = -\dfrac{3}{7}$

Note that -46.4 and 313.6 are not equal angles ... they are coterminal. Same idea holds for -36.9 and 323.1

3. ## Re: further issue with the t substitution

Hi skeeter,

thanks for your patience! Got it now.