Complete square biquadratic

Let $x^4-2^{\tan \lambda}x^2+(\cos\lambda + \cos 2\lambda)x+2^{\tan\lambda-2} = (x^2+ax+b)^2$

Equating coefficients of both sides we get $a = 0, ~~~~b = -2^{\tan \lambda - 1},~~~~ 2ab = 0 = \cos\lambda + \cos 2\lambda, ~~~~~b^2 = 2^{\tan \lambda - 2}$

Solving equations for $b$ we get $\tan \lambda = 0$ i.e $\lambda = n\pi,~ n\in \mathbb{Z}$

But only odd multiples of $\pi$ are permissible as $\cos\lambda+\cos 2\lambda = 0$ .

Hence answer is $\boxed{1000}$