You have,Originally Posted bygarybarlow

Thus,

Thus,

Thus,

Thus,

Thus,

Thus,

Now cheack,

Notice that does not check with the first equation.

However, does.

Thus, you need to solve,

for all

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- Apr 3rd 2006, 01:29 PM #1

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- Apr 3rd 2006, 02:05 PM #2

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- Apr 5th 2006, 10:01 AM #3

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- Apr 5th 2006, 01:08 PM #4

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- Apr 6th 2006, 12:08 AM #5

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- Apr 6th 2006, 02:00 AM #6
## a little trouble!

Well, actually when you are dealing with trigonometric complex functions, a standard trick is to substitute for exponentials. Remember

and after substitution

or

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

so the solution is

or

so that

,

where the logarithm is the usual real one, the principal argument and an integer.

- Apr 6th 2006, 02:47 AM #7

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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 #8

- Apr 6th 2006, 03:10 AM #9

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- Apr 6th 2006, 03:50 AM #10

- Apr 6th 2006, 05:59 AM #11

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