How to solve for when this trigonometric function intersects the line $y=1$? How can I solve for $\alpha$ in
$$4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
on the domain $0\leq\alpha\leq\pi$? Clearly, one solution is when $\alpha=\pi$, but through plotting, it seems to only hold true when $t-r$ is less than a value around $0.3$. When $t-r$ is greater than this value,  it seems to have different solutions.
 A: Excluding the trivial $a=\pi$, reset the problem as
$$(t-r)=\frac{1-\sin\left(\frac{\alpha}{2}\right)}{4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right) }=\frac{1}{4} \left(1-\sin \left(\frac{a}{2}\right)\right) \csc \left(\frac{a}{2}\right)
   \sec ^3\left(\frac{a}{2}\right)\tag 1$$
Let $a=2 \csc ^{-1}(x)$ and $(1)$ becomes
$$(t-r)=\frac{x-1}{4 \left(1-\frac{1}{x^2}\right)^{3/2}}$$
The derivative of the rhs is
$$\frac{x^2 \left(x^2+x-3\right)}{4 (x-1) (x+1)^2 \sqrt{x^2-1}}$$
So, in the given range, the zero of the first derivative corresponds to
$$x^2+x-3=0 \implies x=\frac{\sqrt{13}-1}{2}\implies a=2 \csc ^{-1}\left(\frac{\sqrt{13}-1}{2} \right)$$
At this point, the rhs of $(1)$
$$\frac{\left(\sqrt{13}-3\right) \left(\sqrt{13}-1\right)^3}{8 \left(10-2
   \sqrt{13}\right)^{3/2}} =0.287482$$ which is a minimum value.
So, if $(t-r)$ is greater than this value, there are two roots.
A: Solving for $\alpha$
$$4\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)^{3}\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
$$
4\sin{\left(\frac{\alpha}{2}\right)}\cos{\left(\frac{\alpha}{2}\right)}\left(1-\sin{\left(\frac{\alpha}{2}\right)^2}\right)\left(t-r\right)+\sin{\left(\frac{\alpha}{2}\right)}=1
$$
Let $x = \sin(\frac{\alpha}{2})$
$$
4x\cos(\frac{\alpha}{2})(1-x^2)(t-r)+x=1
$$
Because $\cos(\frac{\alpha}{2})=\sqrt{1-\sin^2(\frac{\alpha}{2}})=\sqrt{1-x^2}$
$$
\Longrightarrow4x\sqrt{1-x^2}(1-x^2)(t-r)+x=1
$$
$$
\Longrightarrow4x(1-x^2)^{\frac{3}{2}}(t-r)+x=1
$$
$$
\Longrightarrow(1-x^2)^3=\frac{\left(1-x\right)^2}{16(t-r)^2x^2}
$$
$$
\Longrightarrow x^2(1+x)^3(1-x)=\frac{1}{16(t-r)^2}
$$
$$
\Longrightarrow  -x^6-2x^5+2x^3+x^2-\frac{1}{16(t-r)^2} = 0
$$
Solving for $\alpha$ amounts to solving for the root of the equation above with dependencies on the value of $t,r$
A: Hint: there are a few ways to proceed in factoring the left hand side. Some ideas:

*

*Factor out $\sin(\alpha/2)$

*Apply double angle formula for $\sin$

*Use Pythagorean identity on $\cos^2(\alpha/2)$. This will allow you to rewrite the left hand side as a polynomial over the variable $\sin(\alpha/2)$.

