general solution of trigonometric equations?? - special cases (1 Viewer)

sazkim · Sep 17, 2016

so far I've learnt that the formulas to the general solutions of trigonometric equations which are:-
sinx = a + k360, (180-a) + k360
cosx = +/-a + k360
tanx = a + k180

but these formulas don't seem to apply in every case
for example, using the formula, the solution to cosx = 0 would be +/-90 + k360
but instead it is 90 + k180

I understand how 90 + k180 is the answer by looking at the cosx graph but does that mean there are special cases that we need to memorise for these general solutions

thank you!!

KingOfActing · Sep 17, 2016

The solution sets are the same.

90 + 360k = 90(4k+1)
-90 + 360k = 90(4k-1)
90 + 180k = 90(2k+1)

Note that in the last equation, if k is even then we get 90(2(2m) +1) = 90(4m + 1) and if k is odd we get 90(2(2m-1)+1) = 90(4m-1)

InteGrand · Sep 17, 2016

sazkim said:
so far I've learnt that the formulas to the general solutions of trigonometric equations which are:-
sinx = a + k360, (180-a) + k360
cosx = +/-a + k360
tanx = a + k180

but these formulas don't seem to apply in every case
for example, using the formula, the solution to cosx = 0 would be +/-90 + k360
but instead it is 90 + k180

I understand how 90 + k180 is the answer by looking at the cosx graph but does that mean there are special cases that we need to memorise for these general solutions

thank you!!

$\noindent The general solution formulas do always apply, it's just that sometimes they can be simplified. For the $\cos x=0$ one, the general formula says the solutions are $90^\circ + 360^\circ k$ or $-90^\circ + 360^\circ k$ for arbitrary integer $k$. This is exactly the same set of solutions as $90^\circ + 180^\circ k$, $k\in \mathbb{Z}$. You can see this is the case by writing out a few of the solutions from the general solutions formula. Or you can write the general solutions here as: $90^\circ + 180^\circ \cdot 2k$ or $90^\circ + 180^\circ \cdot (2k-1)$, $k\in \mathbb{Z}$. As $k$ varies through the integers, the first part of the general solution picks up all the angles that are an even multiple of $180^\circ$ away from $90^\circ$, while the second part gets the ones that differ by odd multiples. So overall we just get angles that are all multiples of $180^\circ$ away from $90^\circ$, so can just write $90^\circ + 180^\circ k$, $k\in \mathbb{Z}$.$

general solution of trigonometric equations?? - special cases (1 Viewer)

sazkim

New Member

KingOfActing

lukewarm mess

InteGrand

Well-Known Member

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