A figure is bounded by the curves $y=\left|\sqrt{2}\sin \frac{\pi x}{4}\right|,y=0,x=2$ and  $x=4$. At what angles to the positive x-axis straight lines must be drawn through $(4,0)$ so that these lines partition the figure into three parts of the same size is/are

Period of  $\left|\sqrt{2}\sin \frac{\pi x}{4}\right|$ is   $\frac{\pi }{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=4$                                 
And range of y is  $[0,\sqrt{2}]$
Since  ${90}^0<\theta$ &  ${\theta }_1<{180}^0$                                         
Any line through $(4,0)$ is $y-0=m(x-4)$ cut $x=2$ at point  $(2,-2m)(\because m<0)$
Area of $\Delta PQR=A_1=\frac{1}{3}\underset{2}{\overset{4}{\int }}\left|\sqrt{2}\sin \frac{\pi x}{4}\right|dx=\frac{1}{2}\times 2\times (-2m)$
$\begin{array}{l}\Rightarrow \frac{1}{3}\sqrt{2}\underset{2}{\overset{4}{\int }}\sin \left(\frac{\pi x}{4}\right)dx=-2m\\ \Rightarrow \frac{\sqrt{2}}{3}\times \frac{4}{\pi }\underset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{\pi }{\int }}\sin xdx=-2m\\ \Rightarrow \frac{4\sqrt{2}}{3\pi }{(-\cos x)}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }=-2m\\ \Rightarrow \frac{4\sqrt{2}}{3\pi }=-2m\Rightarrow -m=\left(\frac{2\sqrt{2}}{3\pi }\right)\\ \therefore {\theta }_1=\pi -{\tan }^{-1}\left(\frac{2\sqrt{2}}{3\pi }\right)\end{array}$

Area of triangle PQS, $A_2= \underset{2}{\overset{4}{\int }}\left|\sqrt{2}\sin \frac{\pi }{4}\right|dx=\frac{1}{2}\times 2\times -2m$

Similarly $\frac{8\sqrt{2}}{3\pi }=-2m\to -m^2=\frac{4\sqrt{2}}{3\pi }$
$\theta =\pi -Ta{n}^{-1}\left(\frac{4\sqrt{2}}{3\pi }\right)$