Setting x  is the are to find. Easy to see that  \(\left(1\right)+\left(2\right)+\left(1\right)=1-\dfrac{\pi}{4}\).

According to the previous post: \(\left(1\right)=1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\).  So that 

             \(\left(2\right)=\left(1-\dfrac{\pi}{4}\right)-2\left(1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\right)=-1+\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{2}\).

Therefore       \(x=\dfrac{\pi}{4}-\left[3.\left(2\right)+2.\left(1\right)\right]=\dfrac{\pi}{4}-\left[\left(-3+\dfrac{\pi}{4}+\dfrac{3\sqrt{3}}{2}\right)+\left(2-\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{3}\right)\right]\)

                                                                   \(=1+\dfrac{\pi}{3}-\sqrt{3}\) .

Setting x  is the are to find. Easy to see that  (1)+(2)+(1)=1−π4.

According to the previous post: (1)=1−√34−π6

.  So that 

             (2)=(1−π4)−2(1−√34−π6)=−1+π12+√32

.

Therefore       x=π4−[3.(2)+2.(1)]=π4−[(−3+π4+3√32)+(2−√32−π3)]

                                                                   =1+π3−√3

 .