I wonder if some one can help me with:
1. to solve the equation
2. to make the function with the e below to loran column
I know it seems simple but for me it is a big help
thank's alot
Solving the equation is pretty trivial isn't it? If $\displaystyle cos^2(z)= -2$ then either $\displaystyle cos(z)= i\sqrt{2}$ or $\displaystyle cos(z)= -i\sqrt{2}$. If your calculator does not allow for inverse trig functions of imaginary numbers (my TI 89 does), use $\displaystyle cos(z)= \frac{e^z+ e^{-z}}{2}= \pm i\sqrt{2}$. Letting $\displaystyle u= e^z$ makes the equation $\displaystyle u^2+ 1= \pm 2 i\sqrt{2} u$ that you can solve with the quadratic equation.
For (2), I simply don't know what you mean by " function with the e below to loran column".