# 4 sin z + cos z = 0

• Apr 3rd 2006, 01:29 PM
garybarlow
4 sin z + cos z = 0
If you could solve the equation in the title (and show me how) I'd be very greatful. Thanks.
• Apr 3rd 2006, 02:05 PM
ThePerfectHacker
Quote:

Originally Posted by garybarlow
If you could solve the equation in the title (and show me how) I'd be very greatful. Thanks.

You have,
$\displaystyle 4\sin z+\cos z=0$
Thus,
$\displaystyle 4\sin z=-\cos z$
Thus,
$\displaystyle 16\sin^2 z=\cos^2 z$
Thus,
$\displaystyle 16\sin^2 z=1-\sin^2 z$
Thus,
$\displaystyle 17\sin^2 z=1$
Thus,
$\displaystyle \sin^2 z=1/17$
Thus,
$\displaystyle \sin z=\pm \sqrt{1/17}$
Now cheack,
Notice that $\displaystyle \sin z=\sqrt{1/17}$ does not check with the first equation.
However, $\displaystyle \sin z=-\sqrt{1/17}$ does.
Thus, you need to solve,
$\displaystyle \sin z=-\sqrt{1/17}$ for all $\displaystyle z$
• Apr 5th 2006, 10:01 AM
mathemagic
4 sin z + cos z = 0

would it not be easier to divide by cos z to get

4 tan z + 1 = 0
and so you'd have to solve tan z = -1/4
that seems easier
• Apr 5th 2006, 01:08 PM
ThePerfectHacker
Quote:

Originally Posted by mathemagic
4 sin z + cos z = 0

would it not be easier to divide by cos z to get

4 tan z + 1 = 0
and so you'd have to solve tan z = -1/4
that seems easier

You first need to show that, $\displaystyle \cos z\not=0$ (which is true), yes it is of course easier.
• Apr 6th 2006, 12:08 AM
mathemagic
$\displaystyle \cos z\not=0$
because whenever cos z is 0, 4 sin z is either 4 or -4, and so the expression would be -4 + 0 or 4 + 0, neither of which are 0, so you can divide by cos z
• Apr 6th 2006, 02:00 AM
Rebesques
a little trouble!
Well, actually when you are dealing with trigonometric complex functions, a standard trick is to substitute for exponentials. Remember
$\displaystyle {\rm e}^{iz}-{\rm e}^{-iz}=2i{\rm sin}z, \ {\rm e}^{iz}+{\rm e}^{-iz}=2{\rm cos}z$

and after substitution
$\displaystyle (1-4i){\rm e}^{iz}+(1+4i){\rm e}^{-iz}=0$ or $\displaystyle {\rm e}^{2iz}=\frac{15-8i}{17}$

(check this please! calculations are not my strong point :()

so the solution is

$\displaystyle 2iz={\rm log}\left(\frac{15-8i}{17}\right)$ or

$\displaystyle z=-\frac{i}{2}{\rm log}\left(\frac{15-8i}{17}\right)$
so that

$\displaystyle z= -\frac{i}{2} \left({\rm log}\left|\frac{15-8i}{17}\right|+i{\rm Arg}\left(\frac{15-8i}{17}\right)+2k\pi i\right)$,

where the logarithm is the usual real one, $\displaystyle {\rm Arg}$ the principal argument and $\displaystyle k$ an integer.
• Apr 6th 2006, 02:47 AM
mathemagic
I've not learnt about the exponential stuff you can do with trigonometric equations like that yet.
I thought of another way to solve it

4 sin z + cos z = 0

let:
4 sin z + cos z = k cos(z - a)

4 sin z + cos z = k cos a cos z + k sin a sin z

k sin a = 4
k cos a = 1
=> tan a = 4
a = 1.33

4 sin z + cos z = k cos(z - 1.33)

k = sqrt(4^2 + 1^2)
k = sqrt(17)

4 sin z + cos z = sqrt(5) cos(z - 1.33)
so now you need to solve
sqrt(17) cos(z - 1.33) = 0
• Apr 6th 2006, 03:01 AM
Rebesques
Quote:

cos(z - 1.33) = 0
I have no idea how this can be solved without exponentials. :(
• Apr 6th 2006, 03:10 AM
mathemagic
to solve sqrt(17)cos (z - 1.33) = 0
you do this:

z- 1.33 = cos -1 (0)
= pi/2, 3pi/2 ...
z = pi/2 + 1.33, 3pi/2 + 1.33 etc
• Apr 6th 2006, 03:50 AM
Rebesques
Oh, :confused: I just realised what :eek: Seems ok!
• Apr 6th 2006, 05:59 AM
ThePerfectHacker
I think we should keep the math simple for the high school students. Not everyone here knows advanced math.