Maximum of $(\sin{x} + \sqrt3)(\cos{x}+1)$ The problem is

Maximum of $(\sin{x} + \sqrt3) (\cos{x}+1)$

I tried to set $\sin(t) = y$, $\cos(t) = x$, so that the problem turned out to be finding
Max $(y + \sqrt3)(x + 1)$
s.t. $x^2 + y^2 =1 $
I tried Lagrange multipliers, but it turns out to be a quartic equation.
I also tried a linear transformation such that the problem becomes
Max $x^2 - y^2$
s.t. $(x + \frac{\sqrt3+1}{2})^2 + (y - \frac{\sqrt3-1}{2})^2 = 1$
Since $x^2-y^2$ is saddle surface, the answer is still not trivial.
Can you help me find the way out?

We can proceed by tangent half angle identities to obtain
$$f(t)=\left(\frac{2t}{1+t^2} + \sqrt3\right)  \left(\frac{1-t^2}{1+t^2} + 1\right) \implies f'(t)=-4\frac{\sqrt 3t^3+3t^2+\sqrt 3t-1}{(1+t^2)^3}$$
and since
$$g(t)=\sqrt 3t^3+3t^2+\sqrt 3t-1 \implies g'(t)=3\sqrt 3t^2+6t+\sqrt 3=3\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^2\ge 0$$
we have that $f'(t)$ has only one root which corresponds to the maximum for $f(t)$.
Now since
$$g'(t)=3\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^2 \implies g(t)=\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^3+c$$
expanding we find $c=-\frac43$ therefore the root is given by
$$\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^3=\frac43 \implies t\approx 0.339$$
