Thread: General solution to a trig equation by turning it into a polynomial equation

by putting $\displaystyle t = tan\frac{\theta}{2}$, find the general solution of the equation

$\displaystyle 3cos\theta + 4sin\theta = 3 - tan\frac{\theta}{2}$


OK i've worked it out to

$\displaystyle t(t-3)(t-3)$

$\displaystyle \therefore t=0, t=3$

but not sure what to do from here.

I also know the general solution for tan is $\displaystyle \theta=\pi n + \alpha$

Originally Posted by djmccabie

by putting $\displaystyle t = tan\frac{\theta}{2}$, find the general solution of the equation

$\displaystyle 3cos\theta + 4sin\theta = 3 - tan\frac{\theta}{2}$


OK i've worked it out to

$\displaystyle t(t-3)(t-3)$

$\displaystyle \therefore t=0, t=3$

but not sure what to do from here.

I also know the general solution for tan is $\displaystyle \theta=\pi n + \alpha$

If $\displaystyle t= tan(\frac{\theta}{2})= 0$ or $\displaystyle t= tan(\frac{\theta}{2})= 3$, what are the possible values of $\displaystyle \theta$?

Well i have done

$\displaystyle arctan (0) =0$

and

$\displaystyle arctan(3) = 1.249045772$ (When calculator set to radians)

but it's not really what i was expecting to see :/ doesn't look right??