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For example, if we have a circle as shown, with angle AOB=120 degrees, and radius 10 units. If we need to find area of the shaded segment, then we find area of sector and subtract area of triangle. Figure Here, area of sector would be $$\frac{120}{360}\pi\times10^2=\frac{100}{3}\pi$$ And the area of the triangle is $$25\sqrt3$$ So the area of segment is $$\frac{100}{3}\pi-25\sqrt3$$ Now, if the angle AOB was a variable $\theta$ and we were told that the shaded area is $\frac{100}{3}\pi-25\sqrt3$. Then we get an equation like $$\frac{\theta}{360}\pi\times100-50\sin\theta=\frac{100}{3}\pi-25\sqrt3$$ How would you solve this? I know that I can equate the terms with $\pi$ in them and get the value of $\theta$, but what if we were given just values instead of something in terms of $\pi$ and $\sqrt3$? (The approximate value is 8.39468514934)

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    $\begingroup$ Note that $\tfrac{120}{360}=\tfrac13$, not $\tfrac19$. $\endgroup$
    – Servaes
    Commented Nov 11, 2020 at 9:28
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    $\begingroup$ You need to be cautious about equating the terms with $\pi$ in them. The example you give has an equation where both sides have been constructed in the same way: the area of a segment minus the area of a triangle (but you have it back to front and an error in division). In this case both terms on each of the two sides of the equation are equal. That is not usually the case, and you cannot then exactly solve an equation containing both $\sin\theta$ and $\theta$. You can only estimate the answer, for example using the answer below, which needs calculus, or using trial and improvement. $\endgroup$
    – WA Don
    Commented Nov 11, 2020 at 9:54
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – user730361
    Commented Nov 11, 2020 at 9:55
  • $\begingroup$ @WADon I think this question can only be solved by graph. $\endgroup$
    – user730361
    Commented Nov 11, 2020 at 9:57
  • $\begingroup$ @Liang : A graph can give one approximation but there are other methods which produce approximations with a very high degree of accuracy. The answer below mentions one (Newton-Raphson). Another is to use a bisection. Suppose you want to solve $\sin\theta + a \theta - b = 0$, find $\theta_1, \theta_2$ which respectively give a negative and positive value (by graph, if you like). Then test the mid point $(\theta_1+\theta_2)/2$. If that is negative replace $\theta_1$, otherwise replace $\theta_2$ and repeat. Each time the gap is halved and you get closer and closer to $0$. $\endgroup$
    – WA Don
    Commented Nov 11, 2020 at 10:21

1 Answer 1

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The equation

$$y=x-\sin(x)$$ is not invertible analytically.

In the range $[0,\pi]$, you have the bracketing

$$\frac{x^3}{\pi^2}\le x-\sin(x)\le\frac{x^2}{\pi}$$ which can give you two gross approximations of $x$. You can extend this range by using symmetries. Then use a few Newton's iterations.

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