What's the size of an angle in a triangle with sides $\sin(x), \cos(x),$ and $\tan(x)$? 
Imagining a scalene triangle with sides $\sin(x), \cos(x)$ and $\tan(x)$, how would you find angle $x$ if it was between $\cos(x)$ and $\sin(x)$ when $0<x<\frac{\pi}{2}$?

I tried using the law of cosines but it lead nowhere and honestly haven't gone very far.
$$\cos(x)=\frac{\sin^2(x)+\cos^2(x)-\tan^2(x)}{2\sin(x)\cos(x)}\\
\cos(x)=\frac{1-\tan^2(x)}{\sin(2x)}$$

From the law of cosines, we have that $$c^2 = a^2 + b^2 - 2ab\cos(C)$$
In this case, we have that $c = \tan(x), a = \cos(x), b = \sin(x)$, and $C = x$. 
Plugging those in, we get $$\tan^2(x) = \cos^2(x) + \sin^2(x) - 2\cos(x)\sin(x)\cos(x)$$
After making the simplification $\cos^2(x) + \sin^2(x) = 1$ and multiplying by $\cos^2(x)$, the result is $$\sin^2(x) = \cos^2(x)-2\cos^4(x)\sin(x)$$
Making the substitution $u = \sin(x)$, we get $$2u^{5}-4u^{3}+2u^{2}+2u-1=0$$
This is a quintic equation with no closed form for the roots. However, WolframAlpha says the relevant root is approximately $0.463$. $x$ is then the $\arcsin$ of this, which means that $$x \approx 0.481$$
