$3\cos \theta  + 4\sin \theta  = 5\,\left[ {\frac{3}{5}\cos \theta  + \frac{4}{5}\sin \theta } \right] = 5\cos (\theta  – \alpha )$

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

अब $3\cos \theta  + 4\sin \theta  = k$

$\therefore$ $5\cos (\theta  – \alpha ) = k $

$\Rightarrow \cos (\theta  – \alpha ) =  \pm 1$

$ \Rightarrow $ $\theta  – \alpha  = {0^o},\,{180^o} $

$\Rightarrow \theta  = \alpha ,\,{\rm{ }}{180^o} + \alpha $.