On squaring both sides of equation (i)
$\begin{array}{rlrl}
& \cos ^2 \theta+16 \sin ^2 \theta-8 \sin \theta \cos \theta & =1 \\
\Rightarrow & & 15 \sin ^2 \theta-8 \sin \theta \cos \theta & =0 \\
\Rightarrow & & \sin \theta(15 \sin \theta-8 \cos \theta) & =0 \\
\Rightarrow & & \sin \theta=0 \text { or } \tan \theta & =\frac{8}{15}
\end{array}$
but $\tan \theta$ is not satisfy the equation (i)
$\begin{aligned}
& \therefore \quad \sin \theta=0 \\
& \Rightarrow \quad \theta=0, \pi \\
& \text { at } \\
& \theta=0 \\
& \sin \theta+4 \cos \theta=0+4=4 \\
& \text { at } \\
& \theta=\pi \\
& \sin \theta+4 \cos \theta=0-4=-4 \\
&
\end{aligned}$