# Thread: Solving Trig equations using the "t" substitution

1. ## Solving Trig equations using the "t" substitution

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?

2. ## Re: Solving Trig equations using the "t" substitution

note ...

$0 \le x < 360^\circ \implies 0\le \dfrac{x}{2} < 180^\circ$

... so when using the $t$ substitution, $\dfrac{x}{2}$ will be calculated only for quads I and II

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

$\dfrac{x}{2} = 180 + \tan^{-1}\left(-\dfrac{1}{3}\right) \approx 161.6^\circ \implies x \approx 323.1^\circ$

$\dfrac{x}{2} = \tan^{-1}\left(\dfrac{5}{3}\right) \implies 0^\circ < \dfrac{x}{2} < 90^\circ$

$\dfrac{x}{2} = \tan^{-1}\left(\dfrac{5}{3}\right) \approx 59.04^\circ \implies x \approx 118.1^\circ$

3. ## Re: Solving Trig equations using the "t" substitution

Hi skeeter,

many thanks once again. It is amazing how many ways there are of getting questions wrong!