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

**djmccabie** · May 18th 2009, 01:31 PM

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

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

OK i've worked it out to

$t(t-3)(t-3)$

$\therefore t=0, t=3$

but not sure what to do from here.

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

**HallsofIvy** · May 18th 2009, 01:49 PM

Originally Posted by djmccabie

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

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

OK i've worked it out to

$t(t-3)(t-3)$

$\therefore t=0, t=3$

but not sure what to do from here.

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

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

**djmccabie** · May 18th 2009, 01:51 PM

Well i have done

$arctan (0) =0$

and

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

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

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