How to solve this equality? [3] $$4x^2 - 6x^4 + \frac{8x^6 - 2x^2 - \frac{1}{x^2}}{16} = 0$$
The equation has a strange look, and as such is probably as it should not be solved. Maybe the roots of trigonometric functions are expressed in terms of angles species $\frac{\pi}{n}$?

$$4x^2-6x^4+\frac{8x^6-2x^2-\frac{1}{x^2}}{16}=0$$
$$4x^2-6x^4+\frac{\frac{8x^8-2x^4-1}{x^2}}{16}=0$$
$$4x^2-6x^4+\frac{8x^8-2x^4-1}{16x^2}=0$$
$$\frac{64x^4-96x^6+8x^8-2x^4-1}{16x^2}=0$$
$$8x^8-96x^6+64x^4-2x^4-1=0$$
