For a warm up here, I recommend you browse the section Wikipedia has on solving the cubic equation using trigonometry. A direct link can be found here. As you mention, the quartic equation can be solved the same way. Further note that one the page for the Quintic Function we have the following quote:
In 1858 Charles Hermite showed that the Bring radical could be characterized in terms of the Jacobi theta functions and their associated elliptic modular functions, using an approach similar to the more familiar approach of solving cubic equations by means of trigonometric functions.
I think the main problem you are having here is that the Abel–Ruffini theorem is often stated as such: "There exists no algebraic solution to the general quintic". This definition leaves a little bit up to the imagination, as the definition of algebraic solution does not mention trigonometric equations. As far as I can tell, the question of whether or not any solution expressible in terms of elementary functions can be expressed in terms of roots is an open question. Take for example the following quote from Dave L. Renfro, taken from a post on Math.SE concerning "polynomials with degree 5 solvable in elementary functions" (which you might find an interesting read):
As of 1999 this was not known -- see Conjecture 2 at the bottom of p. 442 of What is a closed-form number? by Timothy Y. Chow [American Mathematical Monthly 106 #5 (May 1999), 440-448]. I suspect it's still not known, since I believe such a result would have become relatively well known given its intrinsic interest and the fact that understanding the issue at hand doesn't require a lot of advanced mathematical training.
Moreover, here we find a post linking to a couple others, explaining how we can solve the general quintic in terms of the Jacobi Theta Function (which is, in some essence, a trigonometric function) by transforming the general quintic into Brioschi form.
While this is an interesting step, this is not quite what we want. On Wikipedia's page on Closed-form expressions we find the following quote:
Similarly solutions of cubic and quartic (third and fourth degree) equations can be expressed using arithmetic, square roots, and cube roots, or alternatively using arithmetic and trigonometric functions. However, there are quintic equations without closed-form solutions using elementary functions, such as $x^5 − x + 1 = 0$.
This quote makes the claim you desire. Now, how do we prove this? The obvious way is to show that the solutions must be in terms of an antiderivative, as we have known algorithms for determining whether or not an antiderivative can be expressed in terms of elementary functions. In my searchings, I found the following paper by R. Bruce King entitled "Beyond the Quartic Equation", which seems to disprove your conjecture that any polynomials of degree $\geq 5$ can be solved in terms of trigonometric functions. The link is primarily focused on disproving your conjecture for the quintic, but mentions generalizations in a later section (although these pages appear to not be in the free preview. I'll let you know if I find a better link. I would likewise appreciate it if someone else could find such a link).
I hope this provides a comprehensive analysis of your question! If you see any glaring flaws in my answer please share and I will do my best to amend them. Hopefully this should answer your question in as simple a manner as possible!
Note: Admittedly, it's a little hard to get a hold of a good Galois Theory course while in high school, so I don't have a strong enough background to answer with that approach. The only result I use from the field seems to concern the Risch Algorithm, which is used implicitly in the final paper I share.
Yes, we do have a cubic formula! By Cardan's Method...
Cardan's Method: To solve the general cubic$$x^3+ax^2+bx+c=0\tag{i}$$
Remove the $ax^2$ term by substituting $x=\dfrac {y-a}3$. Let the transformed equation be$$y^3+qy+r=0\tag{ii}$$
To solve this depressed cubic, substitute $y=u+v$ to get$$u^3+v^3+(3uv+q)(u+v)+r=0\tag{iii}$$
Put $3uv+q=0$ to get $u=-\dfrac q{3v}$ and substituting this back gives a quadratic in $v^3$. The roots of the quadratic are equal to $u^3,v^3$ respectively. And from our substitution, we get a root as$$y=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}\tag{iv}$$
With the other two roots found with the cube roots of unity.
To find the original root of $(\text i)$, substitute $y$ into your transformation.
Best Answer
Actually this can't be solved by making it a cube.it can't be made because then b and c cannot be arbitrary.there must exist a relation like $ b^2/3.a^2 = c/a $. ; if you differentiate, you will get that the cubic equation having only 1 root has exactly that condition.
This happens because when you want to express it as a cubic, you want to the cube root right!! Then at that point in real plane there is exactly one solution and the other two in complex plane.
So , here we want to transform in terms of other variables but in the actual solution though there is some quadratic things popping out .
Neways we can approximate it by writing the solution as nested radicals(quite similarly to the Methods of S.Ramanujan).
If you have anyproblem to understand this ,please feel free to tell me.I will explain it further.