If x lies in the first quadrant and cos x

Given : $0

Now, $\sin x=\sqrt{1-\cos ^{2} x}=\sqrt{1-\frac{64}{289}}=\frac{15}{17}$

$\mathrm{LHS}=\cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{2 \pi}{3}-x\right)$

$=\cos (30+x)+\cos (45-x)+\cos (120-x)$

$=\cos 30^{\circ} \cos x-\sin 30^{\circ} \sin x+\cos 45^{\circ} \cos x+\sin 45^{\circ} \sin x+\cos 120^{\circ} \cos x+\sin 120^{\circ} \sin x$

$\{$ Using formulas of $\cos (A+B)$ and $\cos (A-B\})$

$=\cos x\left(\cos 30^{\circ}+\cos 45^{\circ}+\cos 120\right)+\sin x\left(-\sin 30^{\circ}+\sin 45^{\circ}+\sin 120^{\circ}\right)$

$=\frac{8}{17}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)+\frac{15}{17}\left(-\frac{1}{2}+\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\right)$

$=\frac{8}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)+\frac{15}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$

$=\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$

= RHS

Hence proved.