solving equation for the variable x

Question

Help to find the value of x :

$e^{\frac{x}{0.52}} = 1.01564 \cos ( tan^{-1} {\frac{x}{0.52}}) $

Thanks for help

Answer 1

You can simplify a little: $\cos(\tan^{-1}(t) = 1/\sqrt{1+t^2}$. But your equation is unlikely to have a closed-form solution. You can find approximate solutions using numerical methods, e.g. Newton's method. Or just ask a CAS, or Wolfram Alpha

Answer 2

As said in comments and answer, numerical methods should be used.

However, you can have good approximations. For simplicity, let us define $t=\frac x {0.52}$ and consider that we look for the zero of equation $$f(t)=\frac{e^t}{\cos(\tan^{-1}(t))}-(1+\epsilon)$$ which, as Robert Israel answered, write $$f(t)=e^t \sqrt{t^2+1}-(1+\epsilon)$$ You can see that $f(0)=-\epsilon$ and expect a solution close to $0$.

Expand as Taylor series to get $$f(t)=1+t+t^2+O\left(t^3\right)-(1+\epsilon)\approx t^2+t-\epsilon$$ So, using the first order you should have $$t_1\approx \epsilon$$ using second order $$t_2\approx\frac{1}{2} \left(\sqrt{1+4 \epsilon }-1\right)$$ If you still wish to avoid numerical methods and stay with approximations, you could use Pade approximants and show that, close to $t=0$, $$e^t \sqrt{t^2+1}\approx\frac{\frac{t^2}{3}-\frac{t}{2}+1}{\frac{5 t^2}{6}-\frac{3 t}{2}+1 }$$ and get $$t_3\approx\frac{9 \epsilon +6-\sqrt{3} \sqrt{-13 \epsilon ^2+12 \epsilon +12}}{2 (5 \epsilon +3)}$$ Using your numbers, the approximate values would then be $$x_1=0.00813280$$ $$x_2=0.00800943$$ $$x_3=0.00800820$$ to be compared to the exact solution Moo gave in a comment.