Ytukyg

How do I solve this equation?

$$sin^2 x +1=2x$$

I have no idea how to attack the problem.

Thanks!

$$1le 1+sin^2(x)le 2implies frac 12le xle 1.$$
the function
$$f:xmapsto sin^2(x)+1-2x$$
is continuous at $[frac 12,1],$

$$f(frac 12)>0; ; f(1)<0,$$
and

$$f'(x)=sin(2x)-2<0.$$

By IVT,
there is a unique solution $alpha$ in $]frac 12,1[$.

$$alpha=lim_{nto+infty}u_n$$

with
$$u_0=1$$
and
$$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}$$

This numerical method is known as Newton-Raphson.

Once it is proved that $f(x)=sin^2(x)-2x+1$ has a unique real zero in the interval $[0,pi/4]$, its numerical determination is simple since $f(x)$ is a positive and convex function on $(0,pi/4)$ (due to $f''(x)>0$), hence by applying Newton's method with starting point $x=0$ we get

$$ rho approx 0.714836$$
in just four steps.

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
$$cos(t) simeqfrac{pi ^2-4t^2}{pi ^2+t^2}qquad (-frac pi 2 leq tleqfrac pi 2)$$ 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 $tapprox 1.428167$ that is to say $xapprox 0.714083$ while the "exact" solution is $xapprox 0.714836$.

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
$$x=frac pi 4 -frac{(pi -3) left(15-6 pi +pi ^2right)}{4 left(12-6 pi +pi ^2right)}approx 0.714837$$

There is no solution in terms of elementary functions - you can solve it only by a numerical algorithm. There is Newton's method - the 'goto' method; but possibly there is an algorithm particular to this one that converges particularly fast. Numerical algorithms is a very highly developed area of mathematics, and for various problems numerical algorithms exist that converge truly astoundingly fast! The one I mentioned - Newtons - converges very fast at nearly every application of it ... and for the vast majority of problems you can do at least that well.