How do I find the angle $\theta$

Question

This is the equation:

$$\sin\theta=0.8\theta$$

Answer 1

Clearly there are three solutions:

Clearly $\theta = 0$ is a solution. By the antisymmetry of the component functions we know there are two symmetric solutions. Simple numerical solution by computer gives the other two: $\theta = \pm 1.13.$

Answer 2

This equation being transcendental, it will not show analytical solutions and numerical methods are required.

You could approximate the solution using the approximation $$\sin(\theta) \simeq \frac{16 (\pi -\theta) \theta}{5 \pi ^2-4 (\pi -\theta) \theta}\qquad (0\leq \theta\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.

Using whole numbers, excluding the trivial solution, you will need to solve for $\theta$ $$4 \theta ^2+(20-4 \pi) \theta -5 \pi (4- \pi)=0$$ giving as approximate acceptable solution $$\theta_\pm=\pm\frac{1}{8} \left(-20+4 \pi +\sqrt{400+160 \pi -64 \pi ^2}\right)\approx \pm 1.12856$$ while the exact solution would be $1.13110$.

Edit for the fun of it !

If, graphing, you notice that the solution is close to $\frac \pi 3$, you could build the $[1,n]$ Padé approximants of the function and get as approximate solutions $$\left( \begin{array}{ccc} n & \theta & \theta \approx \\ 0 & \frac{5}{9} \left(3 \sqrt{3}-\pi \right) & 1.14142 \\ 1 & \frac{5 \left(54+9 \sqrt{3} \pi -8 \pi ^2\right)}{3 \left(93 \sqrt{3}-40 \pi \right)} & 1.13014 \\ 2 & \frac{5 \left(-7533 \sqrt{3}+3861 \pi +96 \sqrt{3} \pi ^2-64 \pi ^3\right)}{3 \left(-16983+3360 \sqrt{3} \pi -320 \pi ^2\right)} & 1.13118 \end{array} \right)$$

Using @b00n heT's idea of solving a biquadratic equation, let us use the $[5,4]$ Padé approximant of $\sin(x)$ built at $x=0$; it is $$\sin(x)=\frac{x-\frac{53 }{396}x^3+\frac{551 }{166320}x^5} {1+\frac{13 }{396}x^2+\frac{5 }{11088} x^4}$$ The difference between this expression and the usual Taylor series is only $\frac{11 x^{11}}{457228800}$.

So, we now need to solve for $x^2$ the biquadratic equation $$491 x^4-26628 x^2+33264=0 \implies x=\pm \sqrt{\frac{6}{491} \left(2219-\sqrt{4470277}\right)}$$ which is $1.13110282$ while the exact solution would be $1.13110259$

Answer 3

Using Taylor's series up to the third term one gets the approximate polynomial equation: $$\theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}=\frac{4}{5}\theta$$ from which we obtain $$\theta\cdot \left(24-20\cdot \theta^2+\theta^4\right)=0$$ which has the solution $\theta=0$ and the two solutions to the biquadratic equation $$\theta^2=\frac{20\pm\sqrt{304}}{2}$$ from which we get $$\theta=\pm\sqrt{\frac{20-\sqrt{304}}{2}}\cong \pm 1.132343637293314$$

Comment: I included only the first three terms in order to get a biquadratic. Otherwise I would have got a polynomial equation of degree 3, which is harder to deal with. Moreover, the approximation works quite well because we know (a priori) that the solution is not too big, as it can be at most $1/0.8=1.25$.