If cos left(x-frac{pi}{3}right), cos x, cos l | vedclass

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(D) If $a, b, c$ are in harmonic progression,then $b = \frac{2ac}{a+c}$.Given $\cos \left(x-\frac{\pi}{3}\right), \cos x, \cos \left(x+\frac{\pi}{3}\r

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$\sqrt{\frac{3}{2}}$

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(D) If $a, b, c$ are in harmonic progression,then $b = \frac{2ac}{a+c}$.Given $\cos \left(x-\frac{\pi}{3}\right), \cos x, \cos \left(x+\frac{\pi}{3}\right)$ are in $H$.$P$.,we have:$\cos x = \frac{2 \cos \left(x-\frac{\pi}{3}\right) \cos \left(x+\frac{\pi}{3}\right)}{\cos \left(x-\frac{\pi}{3}\right) + \cos \left(x+\frac{\pi}{3}\right)}$Using the identity $\cos(A-B)\cos(A+B) = \cos^2 A - \sin^2 B$ and $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$:$\cos x = \frac{2 \left(\cos^2 x - \sin^2 \frac{\pi}{3}\right)}{2 \cos x \cos \frac{\pi}{3}}$$\cos x = \frac{\cos^2 x - \sin^2 \frac{\pi}{3}}{\cos x \cdot \frac{1}{2}}$$\frac{1}{2} \cos^2 x = \cos^2 x - \sin^2 \frac{\pi}{3}$$\sin^2 \frac{\pi}{3} = \cos^2 x - \frac{1}{2} \cos^2 x$$\frac{3}{4} = \frac{1}{2} \cos^2 x$$\cos^2 x = \frac{3}{2}$$\cos x = \pm \sqrt{\frac{3}{2}}$Thus,the correct option is $D$.

If $\cos \left(x-\frac{\pi}{3}\right), \cos x, \cos \left(x+\frac{\pi}{3}\right)$ are in a harmonic progression,then $\cos x=$

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