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How do I solve this equation?

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

I have no idea how to attack the problem.

Thanks!

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  • $\begingroup$ Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods. $\endgroup$ – Blue Nov 21 '18 at 17:01
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    $\begingroup$ Use numerical methods or wolframalpha.com $\endgroup$ – Vasya Nov 21 '18 at 17:01
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    $\begingroup$ This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $\sin^x +1 -2x$ you can see about where it crosses the $x$ axis. $\endgroup$ – Ethan Bolker Nov 21 '18 at 17:02
  • $\begingroup$ @Blue Well, sometimes these equations do have nice solutions e.g. $\sin x=x$. It doesn't seem to be the case here $\endgroup$ – A.Γ. Nov 21 '18 at 17:04
  • $\begingroup$ @AweKumarJha did you mean Aryabhata? $\endgroup$ – Mohammad Zuhair Khan Nov 21 '18 at 17:05
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$$1\le 1+\sin^2(x)\le 2\implies \frac 12\le x\le 1.$$ the function $$f:x\mapsto \sin^2(x)+1-2x$$ 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 $]\frac 12,1[$.

$$\alpha=\lim_{n\to+\infty}u_n$$

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.

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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.

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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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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.

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