By letting $$A=\frac{x}{5}-\frac{\pi}{12}$$ use the sum-to-product formula to simplify the LHS of the equation, I get:
$$\sqrt{2}\cos A-\sqrt{6}\sin A=2\left(\sin \left(A-\frac{7\pi}{12} \right)-\sin \left(3A+\frac{5\pi}{12} \right) \right)$$
$$\sqrt{2}\cos A-\sqrt{6}\sin A=2\left(2\cos \left(2A-\frac{\pi}{12} \right)\sin \left(-A-\frac{\pi}{2} \right) \right)$$
$$\sqrt{2}\cos A-\sqrt{6}\sin A=-4\left(\cos \left(2A-\frac{\pi}{12} \right)\sin \left(A+\frac{\pi}{2} \right) \right)$$
$$\sqrt{2}\cos A-\sqrt{6}\sin A=-4 \left( \cos \left(2A-\frac{\pi}{12}\right)\right)\left( \cos A \right)$$
$$4\cos A \left( \cos \left(2A-\frac{\pi}{12}\right)\right)=\sqrt{6}\sin A- \sqrt{2}\cos A$$
Now, divide the left and right side of the equation by $$\cos A$$ and use the formulas for $$\cos 2A=\frac{1-\tan^2 A}{1-\tan^2 A}$$ and $$\sin 2A=\frac{2\tan A}{1-\tan^2 A}$$ to further simplify the equation yields:
$$\sqrt{6}\tan^3 A+\sqrt{6}\tan^2 A+(2\sqrt{2}-\sqrt{6})\tan A-(2\sqrt{2}+\sqrt{6})=0$$ (*)
It's quite obvious that $$\tan A=1$$ is one of the solution to (*) and use the long division to find the other two roots.
$$(\tan A-1)(\sqrt{6}\tan^2 A+2\sqrt{6}\tan A+2\sqrt{2}+\sqrt{6})=0$$
Since the discriminant of the quadratic expression that we found above is a negative value ($$-8\sqrt{12}$$), we can say the other two roots are imaginary roots.
Hence the solutions are
$$A=\frac{\pi}{4}+2n\pi$$ where n is an integer, i.e.
$$\frac{x}{5}-\frac{\pi}{12}=\frac{\pi}{4}+2n\pi$$ which gives us $$x=\frac{5\pi}{3}+10n\pi$$.