An absolute value integral with a parameter

Case $t\ge0$. The integrand is non-negative, so
$$\int_0^{\pi/2}(\sin x+t\cos x)\,dx=1+t.$$
Setting $1+t=1$ gives $t=0$.

Case $t<0$. Write $t=-k$ with $k>0$. The function $\sin x-k\cos x$ vanishes at $x_0=\arctan k$, is negative before and positive after. Using $\sin x_0=\frac k{\sqrt{1+k^2}}$ and $\cos x_0=\frac1{\sqrt{1+k^2}}$:
$$\int_0^{x_0}(k\cos x-\sin x)\,dx=\bigl[k\sin x+\cos x\bigr]_0^{x_0}=\sqrt{1+k^2}-1,$$
$$\int_{x_0}^{\pi/2}(\sin x-k\cos x)\,dx=\bigl[-\cos x-k\sin x\bigr]_{x_0}^{\pi/2}=\sqrt{1+k^2}-k.$$
The total is $2\sqrt{1+k^2}-k-1$. Setting it equal to $1$:
$$2\sqrt{1+k^2}=k+2\ \Longrightarrow\ 4+4k^2=k^2+4k+4\ \Longrightarrow\ 3k^2=4k.$$
Since $k>0$, $k=\frac43$.

Answer. $t=0$ or $t=-\dfrac43$.