A nice simple trigonometric equation

The first thing that we should do is get rid of the secant and put this problem in terms of tangents. Squaring does the trick.

$\tan^22\theta+2\tan2\theta\tan3\theta+\tan^23\theta=\sec^23\theta=1+\tan^23\theta.$ Simplifying, we have $\tan^22\theta+2\tan2\theta\tan3\theta-1=0.$ Now, we are able to isolate $\tan3\theta.$ $2\tan2\theta\tan3\theta=1-\tan^22\theta,$ so $\tan3\theta=\dfrac{1-\tan^22\theta}{2\tan^2\theta}.$ That right-hand-side looks like a familiar identity; recall that $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}.$ If we flip that around, we realize that our equation is now equivalent to $\tan3\theta=\dfrac{1}{\tan4\theta}$ ! This simplifies to $\tan3\theta\tan4\theta=1.$

Now we are able to make use of the cosine addition formula. Converting to sines and cosines yields $\sin3\theta\sin4\theta=\cos3\theta\cos4\theta,$ or $\cos3\theta\cos4\theta-\sin3\theta\sin4\theta=0.$ This, of course, is equal to $\cos7\theta=0,$ or $\theta=\dfrac{\pi}{14}+\dfrac{\pi n}{7}\forall n\in\mathbb{Z}.$ Thus, $\boxed{N=14}.$

Similar solution as @Vilakshan Gupta 's.

$\begin{aligned} \tan 2 \theta + \tan 3 \theta & = \sec 3 \theta \\ \tan 2 \theta & = \sec 3 \theta - \tan 3 \theta \\ \frac {\sin 2 \theta}{\cos 2 \theta} & = \frac 1{\cos 3 \theta} - \frac {\sin 3 \theta}{\cos 3 \theta} \\ \sin 2 \theta \cos 3 \theta & = \cos 2 \theta - \sin 3 \theta \cos 2 \theta \\ \sin 2 \theta \cos 3 \theta + \sin 3 \theta \cos 2 \theta & = \cos 2 \theta \\ \sin 5 \theta & = \cos 2 \theta \\ \sin 5 \theta & = \sin \left(\frac \pi 2 - 2 \theta \right) \\ \implies 5 \theta & = \frac \pi 2 - 2 \theta \\ 7 \theta & = \frac \pi 2 \\ \implies \theta & = \frac \pi {\boxed{14}} \end{aligned}$

$\tan2\theta+\tan3\theta=\dfrac{\sin2\theta}{\cos2\theta}+\dfrac{\sin3\theta}{\cos3\theta}=\dfrac{\sin5\theta}{\cos2\theta\cos3\theta}=\dfrac{1}{\cos3\theta} \implies \sin5\theta=\cos2\theta=\sin\left(\dfrac{\pi}{2}-2\theta\right)$ and for smallest $\theta$ ,

We must have therefore $5\theta=\dfrac{\pi}{2}-2\theta \implies \theta=\boxed{\frac{\pi}{14}}$ .