Identities?

$\begin{aligned} 2\sin^2 3x - \cos 8x - 1 & = 0 \\ - \cos 8x - \left(1-2\sin^2 3x\right) & = 0 \\ - \cos 8x - \cos 6 x & = 0 \\ \cos 8x & = -\cos 6 x & \small \color{#3D99F6} \text{Using the identity }\cos (\pi - \theta) = - \cos \theta \\ \implies \pi - 8x & = 6x \\ \implies x & = \frac \pi{14} \end{aligned}$

$\implies a + b = 1 + 14 = \boxed{15}$

Note that $2sin^2(3x)-1=cos(6x)$ . Substituting this into the equation nets is $cos (6x)-cos(8x)=0$ . Using the identity $cos(A)+cos(B)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})$ , we can further rewrite our equation as $2cos(7x)cos(2x)=0$ . This means we must have $cos(7x)=0, cos(2x)=0$ or both.

For $x \in (0, \frac{\pi}{2})$ , the minimum solution for $cos(7x)=0$ is $x=\frac{\pi}{14}$ , while the minimum solution for $cos(2x)=0$ is $x=\frac{\pi}{4}$ . The minimum solution overall is thus $x=\frac{\pi}{14}=\frac{a\pi}{b}\Rightarrow a=1, b=15 \Rightarrow a+b=\boxed{16}$

2sin^2x-1 is nothing but -cos6x.......so put value in equation,it will become as Cos8x+cos6x=0....then solve as usual