Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$ Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$ 
$\alpha>0$ & $\beta<\frac{\pi}{2}$
I get the following step after some substitution
$\cos2\alpha + \cos2\beta+\cos2(\alpha+\beta)=-\frac{3}{2}$ from here not able to proceed.

One has $$\cos \alpha\cos\beta(\cos\alpha\cos\beta - \sin\alpha\sin\beta) = -\frac{1}{8}$$
$$1 - \tan\alpha\tan\beta = -\frac{1}{8}(1+\tan^2\alpha)(1+\tan^2\beta)$$
$$8-8\tan\alpha\tan\beta = -1-\tan^2\alpha-\tan^2\beta - \tan^2\alpha\tan^2\beta$$
$$(\tan\alpha\tan\beta-3)^2 + (\tan\alpha-\tan\beta)^2 = 0$$
One then has
$$\tan\alpha=\tan\beta=\sqrt{3}.$$
