# Thread: 4 sin z + cos z = 0

1. ## 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.

2. 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,
$4\sin z+\cos z=0$
Thus,
$4\sin z=-\cos z$
Thus,
$16\sin^2 z=\cos^2 z$
Thus,
$16\sin^2 z=1-\sin^2 z$
Thus,
$17\sin^2 z=1$
Thus,
$\sin^2 z=1/17$
Thus,
$\sin z=\pm \sqrt{1/17}$
Now cheack,
Notice that $\sin z=\sqrt{1/17}$ does not check with the first equation.
However, $\sin z=-\sqrt{1/17}$ does.
Thus, you need to solve,
$\sin z=-\sqrt{1/17}$ for all $z$

3. 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

4. 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, $\cos z\not=0$ (which is true), yes it is of course easier.

5. $
\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

6. ## a little trouble!

Well, actually when you are dealing with trigonometric complex functions, a standard trick is to substitute for exponentials. Remember
$
{\rm e}^{iz}-{\rm e}^{-iz}=2i{\rm sin}z, \ {\rm e}^{iz}+{\rm e}^{-iz}=2{\rm cos}z
$

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

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

so the solution is

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

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

so that

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

where the logarithm is the usual real one, ${\rm Arg}$ the principal argument and $k$ an integer.

7. 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

8. cos(z - 1.33) = 0
I have no idea how this can be solved without exponentials.

9. 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

10. Oh, I just realised what Seems ok!

11. I think we should keep the math simple for the high school students. Not everyone here knows advanced math.