Solving equation that contains cdf and pdf of standard normal distribution I have the following equation: $ x = \frac{1-\Phi(x)}{a\phi(x)}$, where $\Phi$ is the cdf and $\phi$ is the pdf of the standard normal distribution.
How can one solve for $x$? Is there an analytical approach? Or can this only be done numerically?
These might be rather stupid questions (it's quite a while ago that I last was exposed to this), but any suggestions are appreciated. Thanks a lot!

Use the series expansions. You can see here how they can be derived.  Let $a=1$.
The equation is $x\cdot \phi(x)=1-\Phi(x)$. The equation can be multipied by $\sqrt{2\cdot \pi}$. It becomes
$$x\cdot e^{-x^2/2}=\sqrt{2\cdot \pi}-\left(0.5\cdot \sqrt{2\cdot \pi}+\int_0^x e^{-t^2/2} \,dt \right)$$
Using the series expansion the approximated equation is
$$x\cdot \left(1-\frac{x^2}{2}+\frac{x^4}{8}-\frac{x^6}{48}+\frac{x^8}{384}\right)$$ $$=\sqrt{2\cdot \pi}-\left(0.5\cdot \sqrt{2\cdot \pi}+\frac{x}{1\cdot 1}-\frac{x^3}{2\cdot 3}+\frac{x^5}{8\cdot 5}-\frac{x^7}{48\cdot 7}+\frac{x^9}{384\cdot 9}\right)$$
This equation can be solved with Wolfram alpha. The result is $\boxed{x=0.751781}$
To check the result use calculators for the pdf and the cdf. I get 
$$x\cdot \phi(x)=0.751781\cdot 0.30073521=0.226087...\approx 22.609\% $$
$$1-\Phi(x)=1-0.77391=0.22609=22.609\%$$
It looks like that the approximation is fine.
