How to solve $(k-\sqrt{k^2-4y^2})- 10sin^2((m -y)/5)= 0$ and how to embed the solution of equations in other formulas I have a formula like this $5-\sqrt{25 -4 x^2}$, and I know that the value of $x$ is the solution to a rather complicated equation 
$$(k-\sqrt{k^2-4y^2})- 10sin^2((m -y)/5)= 0 $$
The (only) solution to this equation equals: $y= (k=) x$ 
I do not know how to deal with an equation that contains both $x$ and $\sin^2(x)$

1)- Is there a possibility of embedding the equation in the formula so that$5-\sqrt{25 -4* ({\text equation})^2}$
2)- I can use the formula as a function in calculators to find integral, etc..?
  Thanks
3)- How to solve the equation $(k-\sqrt{k^2-4y^2})- 10sin^2((m -y)/5)= 0$?

P.S. please don't downvote my question, I am just a high-school student
 A: I can only answer to your first question.
Let $h(x)$ be a function and let $f(x)=0$ be an equation.
If I understand, your first question asks if it is possible to embed graphically (using mathematical notation) the solution of the equation $f(x)=0$ in the function $h(x)$. $$h({\rm SOLUTIONS}[f(x)=0])$$
Well it is possible. 
We call preimage of $0$ by $f$ the set of $x$'s such that $f(x)=0$ and denote it by $f^{-1}\{0\}$ $$f^{-1}\{0\}:=\{x:f(x)=0\}$$
We then define the image of a set $X$ ($X$ must be a subset of the domain of $h$) by a function $h$ to be the set of $h(x)$'s such that $x\in X$ and denote it by $h[X]$ $$h[X]:=\{h(x):x\in X\}$$
Now we can "insert" the solutions of the equation $f(x)=0$ as an argument of the fucntion $h$
$$h[f^{-1}\{0\}]$$
About the second question: have you tried some softwares like mathematica or asking wolfram alpha?
A: 
I do not know how to deal with an equation that contains both x and $\sin^2(x)$

No one does $\ldots$


How to solve the equation $\Big(k-\sqrt{k^2-4y^2}\Big)-10\sin^2\bigg(\dfrac{m-y}5\bigg)=0$ ?

Let $y=\dfrac k2~\sin t,~$ with $t\in\bigg[0,~\dfrac\pi2\bigg].~$ Then, for $k>0$ we have $$\color{red}t=\arccos\bigg[1-\dfrac{10}k~\sin^2\bigg(\dfrac{2m-k\sin\color{red}t}{10}\bigg)\bigg],$$ and for $k<0,\quad\color{red}t=\arccos\bigg[\dfrac{10}k~\sin^2\bigg(\dfrac{2m-k\sin\color{red}t}{10}\bigg)-1\bigg].~$ Both expressions are recursive 
in nature, so let $t_0=\dfrac\pi4,$ and then iterate using the appropriate above formula.
