be cosx=x^2.

Prove that this equation has exactly two radical .
how I do it?

thanks for help

What do you mean 'prove it has two radicals'?

x=\pm\sqrt{cos(x)}

If you graph this you have parabola. If you want the x intercepts/zeros, note where it crosses the x-axis.

You can also use Newton's method to solve for x if you want to do it algebraically.


f(x)=cos(x)-x^{2}=0

x_{n+1}=\frac{cos(x_{n})+x_{n}(sin(x_{n})+x_{n})}{ sin(x_{n})+2x_{n}}

Perform the iterations and home in on the values.

It is symmetric about the y-axis, so the solution is the positive and negative of the same value. The parabola is concave down with vertex at (0,1).

Other than that, I am sorry but I do not know to what you are referring.