Trigonometry and integer numbers

Note that in general $\tan\left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos(\theta)}{\sin(\theta)}.$

Then $\dfrac{\cos(50^{\circ})}{1 - \sin(50^{\circ})} = \dfrac{\sin(40^{\circ})}{1 - \cos(40^{\circ})} = \dfrac{1}{\tan(20^{\circ})} = \tan(70^{\circ}).$

So we now require that

$\tan(x - 160^{\circ}) = \tan(70^{\circ}) \Longrightarrow x - 160^{\circ} = 70^{\circ} + n*180^{\circ}$

$\Longrightarrow x = 230^{\circ} + n*180^{\circ}.$

The smallest positive value for $x$ then occurs when $n = -1$ and $x = \boxed{50^{\circ}}.$

tan(x-160)=cos50/(1-sin50)

=tan(x-160+360)=cos50/(1-sin50)

=tan(20+x)=cos50/(1-sin50)

=sin(20+x)/cos(20+x)=cos50/(1-sin50)

=sin(20+x)-sin(20+x)sin(50)=cos(50)cos(20+x)

=sin(20+x)=cos(50)cos(20+x)+sin(50)sin(20+x)

Using the identity cos(A-B)=cosAcosB+sinAsinB

=sin(20+x)=cos(20+x-50)

=sin(20+x)=cos(x-30)

=sin(20+x)=sin(90-x+30)

Hence

20+x=120-x

=2x=100

=x=50

First, construct this diagram:

From triangle ABC, we have $\cos 50^\circ = \frac{b}{c}, \sin 50^\circ = \frac{a}{c}$

Hence the RHS expression becomes $\frac{\frac{b}{c}}{1-\frac{a}{c}}=\frac{b}{c-a} = tan\angle AEC$

Since triangle CDE is isosceles, $\angle AEC = \angle DEC = \frac{180^\circ - 40^\circ}{2} = 70^\circ$

Hence $\tan(x-160^\circ) = tan(70^\circ)$

Since $tan(70^\circ) = tan(-110^\circ)$ , smallest integer $x = -110^\circ+160^\circ=\boxed {50^\circ}$