(d) It is given that,
$\cos \left(x-\frac{\pi}{3}\right) \cos x, \cos \left(x+\frac{\pi}{3}\right)$ are in H.P.
$$
\begin{aligned}
& =\frac{2 \cos \left(x-\frac{\pi}{3}\right) \cos \left(x+\frac{\pi}{3}\right)}{\cos \left(x-\frac{\pi}{3}\right)+\cos \left(x+\frac{\pi}{3}\right)} \\
& \Rightarrow \\
& \cos x=\frac{(x)}{\cos \left(x-\frac{\pi}{3}\right)+\cos \left(x+\frac{\pi}{3}\right)} \\
& \Rightarrow \quad \cos x=\frac{2\left(\cos ^2 x-\sin ^2 \frac{\pi}{3}\right)}{2 \cos x \cos \frac{\pi}{3}} \\
& \Rightarrow \quad \cos ^2 x \cos \frac{\pi}{3}=\cos ^2 x-\sin ^2 \frac{\pi}{3} \\
& \Rightarrow \cos ^2 x\left(1-\cos \frac{\pi}{3}\right)=\sin ^2 \frac{\pi}{3} \\
& \Rightarrow \cos ^2 x\left(1-\cos \frac{\pi}{3}\right)=\left(1-\cos ^2 \frac{\pi}{3}\right) \\
& \Rightarrow \cos ^2 x\left(1-\cos \frac{\pi}{3}\right)=\left(1-\cos \frac{\pi}{3}\right)\left(1+\cos \frac{\pi}{3}\right) \\
& \Rightarrow \cos ^2 x=1+\cos \frac{\pi}{3} \\
& \Rightarrow \cos ^2 x=2 \cos ^2 \frac{\pi}{6} \quad \Rightarrow \cos ^2 x=2 \times\left(\frac{\sqrt{3}}{2}\right)^2 \\
& \Rightarrow \cos ^2 x=2 \times \frac{3}{4} \quad \Rightarrow \cos ^2 x=\frac{3}{2} \\
& \Rightarrow \cos x=\sqrt{\frac{3}{2}} . \\
&
\end{aligned}
$$