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

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

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

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

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

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

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