Rapid-fire projectile motion such that all particles land at the same time

I dont know why you took the derivative (plus I think you might have a typo in wolfram alpha), but after you get $t + frac{2u sin theta}{g} = frac{2u sin frac{pi}{4}}{g}$, you can get $theta (t)$ straight from there as:

$$theta (t)=sin^{-1}(sqrt{2}/2-gt/2u)$$

this is a continuous function and passes through $pi /6$, when $t=(sqrt {2}-1)u/g$

I got this solution ?

with:

$$y=v_0,sin(varphi),t-frac{g,t^2}{2}tag 1$$

$Rightarrow$

$$y=0quad,t_0(varphi=varphi_0)=2,{frac {v_{{0}}sin left( varphi _{{0}} right) }{g}}$$

for next bullet: $tmapsto t_0-dt$ and $varphimapsto varphi(t)$ in equation (1)

$$y_t=v_{{0}}sin left( varphi left( t right) right) left( 2,{ frac {v_{{0}}sin left( varphi _{{0}} right) }{g}}-{ dt} right) -frac 1 2,g left( 2,{frac {v_{{0}}sin left( varphi _{{0}} right) }{g}}-{ dt} right) ^{2}tag 2$$

we solve $y_t=0$ and get $dt=(dt_1quad,dt_2)$

$$dt_1=t_0quad dt_2=dt$$ $$dt-2,{frac {v_{{0}} left( -sin left( varphi left( t right) right) +sin left( varphi _{{0}} right) right) }{g}} =0tag 3$$

take the time derivative of equation (3):

$$1-2,{frac {v_{{0}}cos left( varphi left( t right) right) { frac {d}{dt}}varphi left( t right) }{g}} =0$$

solve this equation for $dotvarphiquad Rightarrow$

$${frac {d}{dt}}varphi left( t right) -frac 1 2,{frac {g}{v_{{0}}cos left( varphi left( t right) right) }} =0$$

thus the solution

$$varphi(t)=arcsin left( frac 1 2,{frac {g left( t+{C} right) }{v_{{0}}}} right) quad ,text{with}quad C=t_0$$

$$varphi(t)=arcsin left( frac 1 2,{frac {gt+2,v_{{0}}sin left( varphi _{{0}} right) }{v_{{0}}}} right) tag 4$$

with $varphi(t=1)=varphi_e$ you can obtain the start velocity $v_0$

$$v_0=-frac 1 2,{frac {g}{sin left( varphi _{{0}} right) -sin left( varphi _{{e}} right) }}> 0 ,quad Rightarrowquad varphi_e > varphi_0 $$

I was able to find the issue and apparently (according to meta) its OK to post an answer to my own question.

The working in the question leading to the solution of

$$ alpha(t) = sin^{-1} left (frac{1}{2} + frac{g}{2u}(1-t) right ) $$

is correct, but it's only possible for the first and last bullet to land at the same time for a particular velocity. This was found by using the boundary conditions to verify the first equation, $$ 1 + frac{2u sin frac{pi}{6}}{g} = frac{2u sin frac{pi}{4}}{g} $$

which can be solved for $ u $ to get

$$ u = (1 + sqrt{2})g. $$

So the only actual equation that makes it work, and does so only in the case of this initial speed, is

$$ alpha(t) = sin^{-1} left (frac{1}{2} + frac{1-t}{2(1+sqrt{2})} right ). $$