If cos x=

(b) $\frac{1}{2}$

Given:

$\cos x=\frac{1}{2}\left(a+\frac{1}{a}\right)$

$\cos 3 x=\lambda\left(a^{3}+\frac{1}{a^{3}}\right)$

Now,

$\cos ^{3} x=\frac{1}{8}\left[a^{3}+\frac{1}{a^{3}}+3 a \frac{1}{a}\left(a+\frac{1}{a}\right)\right]$

$\Rightarrow \cos ^{3} x=\frac{1}{8}\left(a^{3}+\frac{1}{a^{3}}+3 \times 2 \cos x\right) \quad\left[\because \cos x=\frac{1}{2}\left(a+\frac{1}{a}\right)\right]$

$\Rightarrow \cos ^{3} x=\frac{1}{8}\left(\frac{\cos 3 x}{\lambda}+6 \cos x\right)$

$\Rightarrow \cos ^{3} x=\frac{1}{8}\left(\frac{4 \cos ^{3} x-3 \cos x}{\lambda}+6 \cos x\right)$

$\Rightarrow \cos ^{3} x=\frac{1}{8}\left(\frac{4 \cos ^{3} x-3 \cos x}{\lambda}+6 \cos x\right)$

$\Rightarrow \cos ^{3} x=\frac{4 \cos ^{3} x}{8 \lambda}-\frac{3 \cos x}{8 \lambda}+\frac{6 \cos x}{8}$

On comparing the powers of $\cos ^{3} x$ on both sides, we get

$1=\frac{4}{8 \lambda}$

$\Rightarrow \lambda=\frac{1}{2}$