You have,Originally Posted by garybarlow
Notice that does not check with the first equation.
Thus, you need to solve,
Well, actually when you are dealing with trigonometric complex functions, a standard trick is to substitute for exponentials. Remember
and after substitution
(check this please! calculations are not my strong point )
so the solution is
where the logarithm is the usual real one, the principal argument and an integer.
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
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