$\cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)=$

$x-\tan ^{-1} \frac{4}{3}$

$\cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)$        .....(1)    

we know that $\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1$

let $\frac{3}{5}=\cos \theta \quad \frac{4}{5}=\sin \theta$

so $\tan \theta=4 / 3 \Rightarrow \theta=\tan ^{-1}(4 / 3)$    .......(2)

Substituting in (1)

$\cos ^{-1}(\cos \theta \cos x+\sin \theta \sin x)$

$\cos ^{-1} \cos (x-\theta)$

$=x-\theta$

We know that $\cos A \cos B+\sin A \sin B$

$=\cos (A-B)$

$\cos^{-1} \cos (A)=A$

from (2) substituting $\theta$

we get $x-\tan ^{-1} 4 / 3$