How do dis?

$x=\dfrac{\pi}{4}\\ \therefore y = \dfrac{1}{\sin\left(2\cdot\dfrac{\pi}{4}\right)}=1\\ \dfrac{dy}{dx}=(\csc(2x))'=-2\csc(2x)\cot(2x)\\ \dfrac{dy}{dx}|_{x=\frac{\pi}{4}}=-2(1)(0)=0\\ y-1=0(x-\dfrac{\pi}{4})\\ y=1\\ \therefore\text{The equation of the normal to }y=\dfrac{1}{\sin(2x)}\text{ when }x=\pi/4\text{ is }\boxed{y=1}$

I think this is the first time I do an application question :P

Find the equation of the normal to y = 1/sin(2x) at the point where x = pi/4

We can write this as

y = [sin(2x) ]^-1 

When x  = pi/4, y =  1 / sin (2 (pi/4))  =  1 / sin (pi/2)  = 1

y '  =   -2 [ sin (2x)]-2 * [ cos(2x)]  =   [-2 cos (2x)] / [ sin (2x)] ²  =   -2cot (2x)csc(2x)

At pi/4....the slope of the tangent line  =   -2 [ cot (2(pi/4) ) csc (2(pi/4) )=

-2 [ cot (pi/2)] [csc(pi/2)]  =  0

So the equation of the tangent line  is   ...    y - 1  = 0 ( x - pi/4)  →  y  = 1

So...the  slope of a line normal to this will be undefined.....and the equation of the line will be  x  = pi/4

Here's a graph to show this : https://www.desmos.com/calculator/qyxqzuupbs