<s>[INST] Solving $\sin^2 x +1=2x$ How do I solve this equation?
$$\sin^2 x +1=2x$$ 
I have no idea how to attack the problem.  
Thanks!  
 [/INST] Just for the fun of the approximation.
Using the double angle formula Rewrite the equation as $$\cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
$$\cos(t)+2t=3$$
Now, using the approximation 
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad (-\frac \pi 2 \leq t\leq\frac \pi 2)$$ we get the cubic equation
$$2 t^3-7 t^2+2 \pi ^2 t-2 \pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $t\approx 1.428167$ that is to say $x\approx 0.714083$ while the "exact" solution is $x\approx 0.714836$.
We can even do better building the $[2,2]$ Padé approximant around $x=\frac \pi 4$
$$\cos(2x)+4x-3=\frac{(\pi -3)+2 \left(x-\frac{\pi }{4}\right)+\left(2-\frac{2 \pi }{3}\right)
   \left(x-\frac{\pi }{4}\right)^2 } {1-\frac{2}{3} \left(x-\frac{\pi }{4}\right)^2 }$$ Solving the quadratic
$$x=\frac \pi 4+\frac{6-\sqrt{252-144 \pi +24 \pi ^2}}{4 \pi -12}\approx 0.714837$$
To even avoid solving the quadratic equation,  building instead the $[1,3]$ Padé approximant around $x=\frac \pi 4$ would lead to
$$x=\frac \pi 4 -\frac{(\pi -3) \left(15-6 \pi +\pi ^2\right)}{4 \left(12-6 \pi +\pi ^2\right)}\approx 0.714837$$
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