Stationary point for curve y=3sec(2x)

I recently learnt gradients to non-algebraic functions and now realising some are available on the formula sheet.

Anyway, so if I know the gradient is

F’(x)=6sec(2x)tan(2x) and stationary point being f’(x)=0

Is there a shortcut to find the value of x? I have tried to use the double angle identities with no fruition as I’m left with both sin and cos still in the function but I might be over looking something.

The gradient should be 2sec(2x)tan(2x). Not important for stationary points as its when its = 0 but ... So just think about factors so when sec(2x) or tan(2x) are zero as their product must be zero.

If unsure, a sketch of sec(2x) might be helpful for thinking about it.

No option for uploading photos at the moment after the recent "upgrade". Youd need to upload elsewhere and link with the picture icon/button.

Apologise, I missed the 3 from the original function.

Well that sucks about the photo issue.

ah, so without recalling the shape of the sec(x) graph, it’s sorta impossible to answer.

If it was 3sec(2x) then the derivative is correct. To find the stationary points you could note sec=1/cos so this corresponds to
sec(2x)tan(2x) = sin(2x)/cos^2(2x) = 0
and you can argue about either expression being zero to get the stationary points. However, you should be able to sketch sec, even if you have to use the 1/cos definition

Let’s say I could recalled sec x, then sec(2x)=0 would be invalid as it doesn’t cross the x-axis..

Tan(x)= x+π

Tan(2x) would suggest a stretch in x… okay, so if 0<x<π
Then new range would be 0<x<2π

Am I on the right path?

So, the of chance I can recall sec 2x, I can take the long path using identities and consider that sin x = x+π

Hence, x=0,π and 2π

So,

2x=0
2x=π
2x= 2π

So, x=2/π can be the only permissible x for the one point it’s looking for.

Then inputting that into the original equation tells us y=-3