Question

Solve:

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

Answer 1

Given,

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

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

Let

$\frac{3}{5}=\cos \alpha ,\frac{4}{5}=\sin \alpha$

Therefore,

$={\cos }^{-1}(\cos \alpha \cos x-\sin \alpha \sin x)$

$={\cos }^{-1}\left(\cos \right(x+\alpha \left)\right)$

$=x+\alpha$ where $\alpha ={\cos }^{-1}\frac{3}{5}$

Hence, this is the answer.