The bigger they are,the harder they fall

We know that $cos(x) \le 1$ Hence the LHS $2cos^{2}(x^8+6x^{7}+67x^{6}+45x^{5}+89x^{2}+23x) \le 2$

But the RHS of the equation is $13^{x}+13^{-x}$ ,clearly both the terms are positive. So by applying AM-GM inequality we get $13^{x}+13^{-x} \ge 2$

So LHS is less than equal to 2 and RHS is greater than or equal to 2.So the only possibility for equality is LHS=RHS = 2

For the RHS of the expression to be equal to 2, we must have the equality case of AM-GM ,that is both the terms are equal ,hence $13^{x} = 13^{-x}$ $13^{2x} = 1$ $x = 0$

We can see that $x = 0$ readily satisfies the LHS and makes it equal to $2$ .

So there is only $\boxed{1}$ possible solution to this equation.