Solve for $\theta$: $x = \theta - \sin\theta$

Question

Solve for $\theta$:

$$x = \theta - \sin\theta$$

Is this type of isolation a matter of identities? If so, which one(s)?

Answer 1

As said in comments, you need some numerical methods for this kind of transcendental equations.

How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 \leq \theta \leq \pi$, the maginficent approximation $$\sin(\theta) \simeq \frac{16 (\pi -\theta) \theta}{5 \pi ^2-4 (\pi -\theta) \theta}\qquad \text{for} \qquad 0\leq \theta\leq\pi$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation $$4\, \theta ^3-4 (x+\pi -4)\,\theta ^2+ \pi\left(4 x+5 \pi -16 \right)\,\theta-5 \pi ^2 x=0$$ which can be solved.

In the table below, I produced some values for the solution. $$\left( \begin{array}{ccc} x & \text{approximation} & \text{exact} \\ 0.0 & 0.00000 & 0.00000 \\ 0.1 & 0.84987 & 0.85375 \\ 0.2 & 1.08166 & 1.08369 \\ 0.3 & 1.24765 & 1.24852 \\ 0.4 & 1.38201 & 1.38228 \\ 0.5 & 1.49726 & 1.49730 \\ 0.6 & 1.59958 & 1.59959 \\ 0.7 & 1.69251 & 1.69259 \\ 0.8 & 1.77828 & 1.77851 \\ 0.9 & 1.85843 & 1.85881 \\ 1.0 & 1.93404 & 1.93456 \\ 1.1 & 2.00590 & 2.00655 \\ 1.2 & 2.07463 & 2.07538 \\ 1.3 & 2.14070 & 2.14151 \\ 1.4 & 2.20451 & 2.20534 \\ 1.5 & 2.26636 & 2.26717 \\ 1.6 & 2.32651 & 2.32726 \\ 1.7 & 2.38517 & 2.38584 \\ 1.8 & 2.44254 & 2.44308 \\ 1.9 & 2.49876 & 2.49915 \\ 2.0 & 2.55398 & 2.55420 \end{array} \right)$$

Another possible way would be Taylor expansion and series reversion to get $$\theta=t+\frac{t^3}{60}+\frac{t^5}{1400}+\frac{t^7}{25200}+\frac{43 t^9}{17248000}+\frac{1213 t^{11}}{7207200000}+\frac{151439 t^{13}}{12713500800000}$$ where $t=\sqrt[3]{6x}$