Solve the following equation : $2(2(x^{2}-2)^{2}-7)^{2}-1=\sqrt{1-x^2}$ Solve in $R$ the following equation : 
$2(2(x^{2}-2)^{2}-7)^{2}-1=\sqrt{1-x^2}$
$x=0$ clearly one  solution , but how I find other solutions 
How many solutions?
My attempt is put 
$ y =\sqrt{1-x ^ 2} $ 
But I do not know how to solve  this type of equations?
 A: You must take care that squaring processes introduce extra roots.
In the present case, using the symmetry, I should simply plot the function
$$f(x)=2(2(x^{2}-2)^{2}-7)^{2}-1-\sqrt{1-x^2}$$ for $0 \leq x \leq 1$ and visually notice that there is a root close to $x=0.5$.
To approximate the root, use a Taylor expansion around this point to get
$$f(x)=\left(\frac{17}{32}-\frac{\sqrt{3}}{2}\right)+\left(\frac{49}{2}+\frac{1}{\sqrt{3}}
   \right) \left(x-\frac{1}{2}\right)+O\left(\left(x-\frac{1}{2}\right)^2\right)$$ Ignoring the higher order terms, this gives as an estimate
$$x=\frac 12+\frac{3 \left(16 \sqrt{3}-17\right)}{16 \left(147+2 \sqrt{3}\right)}\approx 0.513350$$ while the  solution is $\approx0.512603$.
Added for your curiosity
We could get a better approximation if, instead of using Taylor expansion, we use the $[1,1]$ Padé approximant. This would give as an estimate
$$x=\frac 12+\frac{6954 \sqrt{3}-7209}{310977+42536 \sqrt{3}}\approx 0.512572$$
A: $X=\sqrt{1-x^2}$, replacing you have the equation $8X^8+32X^6-8X^4-80X^2-X+49=0$. It has $X=1$ and another positive ($<1$) solution.
A: Expanding the left-hand side we get
$$8\,{x}^{8}-64\,{x}^{6}+136\,{x}^{4}-32\,{x}^{2}+1$$
squaring this term and subtracting $(1-x^2)$
$${x}^{2} \left( 64\,{x}^{14}-1024\,{x}^{12}+6272\,{x}^{10}-17920\,{x}^{
8}+22608\,{x}^{6}-8832\,{x}^{4}+1296\,{x}^{2}-63 \right) 
=0$$
by a numerical method we get the following solutions
$$-.5126032239, -.3954020655, -.3272996816, 0., 0., .3272996816, .3954020655, .5126032239$$
