$\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$

$2 \cos (\beta-\gamma)+2 \cos (\gamma-\alpha)+2 \cos (\alpha-\beta)=-3$

$1+1+1+2(\cos \beta \cos \gamma+\sin \beta \sin \gamma)+2(\cos \gamma \cos \alpha+\sin \gamma \sin \alpha)+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)$

$=0$

$\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+\left(\sin ^{2} \gamma+\cos ^{2} \gamma\right)+2 \cos \alpha \cos \beta+2 \cos \beta \cos \gamma+2 \cos$

$\gamma \cos \alpha$

$+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma+2 \sin \gamma \sin \alpha=0$

$\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma\right.$

$+2 \sin \gamma \sin \alpha)+\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right.$

$+2 \cos \alpha \cos \beta+\cos \beta \cos \gamma+\cos \gamma \cos \alpha)=0$

$(\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}=0$

Only Possible when

$\sin \alpha+\sin \beta+\sin \gamma=0$

$\cos \alpha+\cos \beta+\cos \gamma=0$

$\sin \theta+\sin 4 \theta+\sin 7 \theta=0$

$2 \sin \left(\frac{\theta+7 \theta}{2}\right) \cos \left(\frac{7 \theta-\theta}{2}\right)+\sin 4 \theta=0$

$\Rightarrow \sin 4 \theta[2 \cos 3 \theta+1]=0$

$\Rightarrow \sin 4 \theta=0 \Rightarrow 4 \theta=0, \pi, 2 \pi, 3 \pi, 4 \pi$

$\Rightarrow \theta=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi$

but 0 and $\pi$ are not included.

and $2 \cos 3 \theta+1=0 \Rightarrow \cos 3 \theta=\frac{-1}{2}$

$\Rightarrow 3 \theta=\frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3} \quad \Rightarrow \quad \theta=\frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}, \frac{10 \pi}{9}$

but $\frac{10 \pi}{9} \notin(0, \pi)$

So, $\theta=\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}$

$2 \cos 2 x \cos x+2 \cos 3 x \cos x=0$

$\Rightarrow 2 \cos x(\cos 2 x+\cos 3 x)=0$

$2 \cos x 2 \cos 5 x / 2 \cos x / 2=0$

$x=\frac{\pi}{2}, \frac{3 \pi}{2}, \pi, \frac{\pi}{5}, \frac{3 \pi}{5}, \frac{7 \pi}{5}, \frac{9 \pi}{5}$

7 Solutions

$8 \cos x\left(\cos ^{2} \frac{\pi}{6}-\sin ^{2} x-\frac{1}{2}\right)=1$

$\Rightarrow 8 \cos x\left(\frac{1}{4}-\left(1-\cos ^{2} x\right)\right)=1$

$\Rightarrow 8 \cos x\left(\cos ^{2} x-\frac{3}{4}\right)=1$

$\Rightarrow 2 \cos 3 x=1 \Rightarrow \cos 3 x=\frac{1}{2}$

$\therefore 3 x+2 n \pi \pm \frac{\pi}{3}, n \in I$

$\Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}$

$\ln \mathrm{x} \in[0, \pi]: \mathrm{x}=\frac{\pi}{9}, \frac{2 \pi}{3}+\frac{\pi}{9}, \frac{2 \pi}{3}-\frac{\pi}{9}$ only

$\operatorname{sum}=\frac{13 \pi}{9}$