The maximum value of 3 cos ⁡ θ + 5 sin ⁡ θ - π 6 for any real value of θ is :

Given $f\left(\theta \right)=3\cos \theta +5\sin \left(\theta -\frac{\pi }{6}\right)$

$\Rightarrow f\left(\theta \right)=3\cos \theta +5\left(\sin \theta ·\frac{\sqrt{3}}{2}-\cos \theta ·\frac{1}{2}\right)$

$\Rightarrow f\left(\theta \right)=\frac{5\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta$

Now using the concept $a\cos x+b\sin x+c \in \left[c-\sqrt{a^2+b^2}, c+\sqrt{a^2+b^2}\right]$, we can write

Maximum value of $f\left(\theta \right)$ is $\sqrt{{\left(\frac{5\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{75}{4}+\frac{1}{4}}=\sqrt{19}$