product of angle and their cosecant. if angle in degree

Question

$$\alpha\csc(\alpha)= \text{a constant}$$ Find the value of $\alpha$, if $\alpha$ is in degrees.

Example; $$\alpha\csc\left(\frac\alpha2\right) = 120^0$$ Find the value $\alpha$ ; without assuming the $\alpha$ value

Answer 1

Let me make your quizz a problem. You are looking for the solution of $$\alpha\csc\big(\frac \alpha k\big)= a$$ where $a$ and $k$ are known constants.

For simplification, let $x=\alpha k$ which makes the equation $$kx\csc(x)=a$$ You could already notice that if $x$ is a solution, $-x$ is another one.

Since we look for the solution, we can say that we look for the zero of function$$f(x)=\beta x-\sin(x)$$ using $\beta=\frac k a$. Beside the trivial solution $x=0$, you can notice that there is no other solution if $\beta \geq 1$.

You could be interested by this question of mine. we can find very good approximations of the solution using the magnificent formula $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than 1,400 years ago).

Let us use it for that case you gave : $k=2$, $a=\frac {2\pi}3$ which makes $\beta=\frac 3 \pi$. Reducing to same denominator gives $$f(x)\approx\frac{ (\pi -6 x) x (2 x+\pi )}{5 \pi ^2-4 (\pi -x) x}$$ which, in the definition interval, only cancels for $x=\frac \pi 6$ which, by chance, happens to be the exact solution of the problem.

Just for illustration purposes, let us consider $a=\frac {\pi}2$ which, for $k=2$, makes $\beta=\frac 8{3 \pi}$. Replacing would give $$f(x)\approx \frac 8{3\pi}\frac{ x \left(-4 x^2-2 \pi x+\pi ^2\right)}{5 \pi ^2-4 (\pi -x) x}$$ and solving the quadratic will give as an approximate solution $$x=\frac{1}{4} \left(\sqrt{5}-1\right) \pi\approx 0.970806$$ while the exact solution, which would require numerical methods (such as Newton), would be $\approx 0.975334$