$t$ as in tangent?

Let the integral be $I(t)$ . For $t > 0$ ,

$\begin{aligned} I(t \ge 0) & = \int_0^\frac \pi 2 (\sin x + t \cos x) \ dx \\ & = - cos x + t \cos x \ \bigg|_0^\frac \pi 2 \\ & = 1 + t \end{aligned}$

There is no solution for $t > 0$ .

For $t \le 0$ ,

$\begin{aligned} I(t\le 0) & = \int_0^\frac \pi 2 |\sin x + t \cos x| \ dx \\ & = \int_0^\frac \pi 2 \left|\sqrt{1+t^2} \sin \left(x + \tan^{-1} t \right) \right| \ dx \\ & = \sqrt{1+t^2} \int_0^\frac \pi 2 \left|\sin \left(x + \tan^{-1} t \right) \right| \ dx \\ & = \sqrt{1+t^2} \left(-\int_0^{-\tan^{-1}t} \sin \left(x + \tan^{-1} t \right) dx + \int_{-\tan^{-1}t}^\frac \pi 2 \sin \left(x + \tan^{-1} t \right) dx \right) \\ & = \sqrt{1+t^2} \left(\cos \left(x + \tan^{-1} t \right) \bigg|_0^{-\tan^{-1} t} + \cos \left(x + \tan^{-1} t \right) \bigg|_\frac \pi 2^{-\tan^{-1} t} \right) \\ & = \sqrt{1+t^2} \left(1 - \frac 1{\sqrt{1+t^2}} + 1 + \frac t{\sqrt{1+t^2}} \right) \\ & = 2\sqrt{1+t^2} - 1 + t \end{aligned}$

For $I(t\le 0) = 1$ ,

$\begin{aligned} \implies 2 \sqrt{1+t^2} - 1 + t & = 1 \\ 4 + 4t^2 & = 4-4t + t^2 \\ t(3t+4) & = 0 \\ \implies t & = \begin{cases} 0 \\ - \frac 43 \end{cases} \end{aligned}$

The sum of values of $t$ satisfying the equation is $0 - \dfrac 43 \approx \boxed{-1.33}$ .