I recommend to draw those function (for $t = 1,3,-5 \dots$). i) $\forall t\geq 0$ For $x \in \left[0, \frac{\pi}{2} \right]$ we have $\sin x +t \cos x \geq 0\ $ and so $$\int_{0}^{\frac{\pi}{2}}( \sin x+t\cos x) \ {\rm d}x = t+1 \implies \boxed{t =0}$$ ii) $\forall t< 0$, put $-t = k$ and so $k>0$ We need to solve $\sin x =k\cos x$ and so $x = \arctan k$ and then we have \begin{align*} 1 &=-\int_{0}^{\arctan k}( \sin x-k\cos x) \ {\rm d}x + \int_{\arctan k}^{\frac{\pi}{2}}( \sin x-k\cos x) \ {\rm d}x =\\ & = -\left[ -\cos x - k\sin x \right]_0^{\arctan k} +\left[ -\cos x - k\sin x \right]_{\arctan k}^{\frac{\pi}{2}} \\ & = -\left[ -\cos (\arctan k) - k\sin (\arctan k)+1 \right]_0^{\arctan k} +\left[ - k +\cos (\arctan k) + k\sin (\arctan k)\right]_{\arctan k}^{\frac{\pi}{2}} \\ & = -k-1+\frac{2}{\sqrt{k^2+1}}+\frac{2k^2}{\sqrt{k^2+1}}\ \text{(note 1)} \\ & = -k-1+2\sqrt{k^2+1} \\ 4+4k+k^2 & = 4k^2+4 \\ k_{0,1} &= \left \{0, \frac{4}{3} \right \} \implies \boxed{t = -\frac{4}{3}} \end{align*} Note 1 (it is not so hard to find it): $$ \sin (\arctan x) = \frac{x}{\sqrt{x^2+1}} \ , \ \cos (\arctan x) = \frac{1}{\sqrt{x^2+1}} $$