Root of the given polynomial $4x^3-3x-p$ Given the equation $$ 4x^3 - 3x-p=0 $$
In the question we were required to find the root of this equation in the interval $[1/2,1]$ and $-1\le p\le 1$. The answer is arrived at by substituting x as $\cos(\theta)$
and using the multiple angle formula. However assuming I am not able to think of this substitution in the exam(it does not appear to be a very intuitive one considering the range of $x$ is not $[-1,1]$), is there another method to arrive at this answer?
 A: You face a depressed cubic equation
$$4x^3-3x-r=0$$ I changed the notation for obvious reasons if you look here. Let us exclude the cases $r=\pm 1$ for which the roots are obvious.
So, following the method, we have $\Delta=432 \left(1-r^2\right)$ which is always positive; then three real roots. So, using the trigonomatric method (!!) for three real roots, we end with
$$x_k=\cos \left(\frac{2 \pi  k}{3}-\frac{1}{3} \cos ^{-1}(r)\right)\qquad \text{with}\qquad k=0,1,2$$
A: If the solution is not rational, you can use Cardano's method to find one real root. $4x^3-3x-p=0$ is equivalent to $x^3-\frac{3}{4}x-\frac{p}{4}=0$. And now you let $t=u+v$, which only holds if $-\frac{3}{4}=-3uv$ and $-\frac{p}{4}=-(u^3+v^3)$. Therefore, $u=\frac{1}{4v}$ and $\frac{p}{4}=v^3+\frac{1}{64v^3}$. Multiply everything by $v^3$ and you get $v^6-\frac{p}{4}v^3+\frac{1}{64}=0$, which is a triquadratic equation and can be solved by the Bhaskara's formula. $u^3=\frac{p}{8}\pm\sqrt \frac{p^2-1}{64}$, so $u=\sqrt[3]{\frac{p}{8}\pm\sqrt \frac{p^2-1}{64}}$. $v$ will be its conjugated, so $t=\sqrt[3]{\frac{p}{8}+\sqrt \frac{p^2-1}{64}}+\sqrt[3]{\frac{p}{8}-\sqrt \frac{p^2-1}{64}}$.
