If $x^2+\frac{1}{2x}=\cos \theta$, evaluate $x^6+\frac{1}{2x^3}$. If $x^2+\frac{1}{2x}=\cos \theta$, then find the value of $x^6+\frac{1}{2x^3}$.
If we cube both sides, then we get $x^6+\frac{1}{8x^3}+\frac{3x}{2} \cdot \cos \theta=\cos ^3 \theta$ but how can we use it to deduce required value?
 A: One can get a rather messy solution by using the solution for the depressed cubic, $x^3 + px + q = 0$. Here $p = -\cos(\theta), q = 1/2$.
https://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic 
The solution involves a trig substitution $x = u \cos(\alpha)$, and a trig identity is used to reduce the problem to a tractable form. Using the identity $$4 \cos^3(\alpha) - 3 \cos(\alpha) -\cos(3\alpha) = 0$$
and choosing a nice value for $u$ we get a solution for $x$ in terms of $\theta$. Then it's a matter of plugging in values for $x$.
A: $$2x^3+1=2x\cos(a)\implies x^3-x\cos(a)+{1\over 2}=0$$Let $x=y+{b\over y}\implies y={1\over 2}(x+\sqrt{x^2-4b})$ Using back substitution$$\left(y+{b\over y}\right)^3-\left(y+{b\over y}\right)\cos(a)+{1\over 2}=0$$$$\implies {1\over 2}y^3+y^6+b^3+y^4(3b-\cos(a))+y^2(3b^2-b\cos(a))=0$$
Now substitute $b={\cos(a)\over 3}$ and $z=y^3$ $$z^2+{z\over 2}+{\cos^3(a)\over 27}=0\implies z={1\over 36}\left(\sqrt{3}\sqrt{27-16\cos^3(a)}-9\right)$$
Now substitute back $z=y^3$ and it will give the $3$ values of $y$.
Once you get the values of $y$ substitute back the values of $x$ and you will get $3$ quadratics  of $x$ 
BINGO! 
A: Let $c = \cos \theta$.  You have $x^3 = c x - 1/2$, so $x^6 = (c x - 1/2)^2$ while $1/x = 2 c - 2 x^2$.  Then I get
$$\eqalign{x^6 + \dfrac{1}{2x^3} &= - 3 c^2 x^2 - 3 c x + 4 c^3 - \frac{3}{4}\cr & =
   -\dfrac{3}{2} (1 + \cos(2\theta)) x^2 - 3 \cos(\theta) x + \cos(3\theta) + 3 \cos(\theta) - \dfrac{3}{4}}$$
