$\text{ Let the function }f:(0,\pi )\to \mathrm{ℝ}\text{ be defined by f }\left(\theta \right)=(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^4$

$\text{ Suppose the function }f\text{ has a local minimum at }\theta \text{ precisely when }\theta \in \left\{{\lambda }_1\pi ,\dots ,{\lambda }_{\mathrm{r}}\pi \right\}$

$\text{ where }0<{\lambda }_1<\cdots <{\lambda }_r<1\text{ . Then the value of }{\lambda }_1+\cdots +{\lambda }_r\text{ is }$

$\begin{array}{c}f\left(\theta \right)=(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^4\\ = 1+\sin 2\theta + {\left(1-\sin 2\theta \right)}^2\\ f\left(\theta \right)={\sin }^{2}2\theta -\sin 2\theta +2\\ f^{\mathrm{\prime }}\left(\theta \right)=2(\sin 2\theta )\cdot (2\cos 2\theta )-2\cos 2\theta \\ =2\cos 2\theta (2\sin 2\theta -1)\end{array}$

$$\begin{gathered} f^{\text{'}}\left(\theta \right)=0 \Rightarrow \sin 2\theta =\frac{1}{2} or \cos 2\theta =0 \\ 2\theta =\frac{\pi }{6}, \frac{5\pi }{6} or 2\theta =\frac{\pi }{2}, \frac{3\pi }{2} \end{gathered}$$

critical points

$\begin{array}{r}\text{ so, minimum at }\theta =\frac{\pi }{12},\frac{5\pi }{12}\\ {\lambda }_1+{\lambda }_2=\frac{1}{12}+\frac{5}{12}=\frac{6}{12}=\frac{1}{2}\end{array}$