The minimum value of 4 cos x – 3 sin x + 7 is

$4 \cos x-3 \sin x+7$

Express $4 \cos x-3 \sin x$ as $a \cos (x+A)$

i. e. $4 \cos x-3 \sin x=a[\cos x \cos A-\sin x \sin A]$

on compairing coefficients of $\sin x$ and $\cos x$,

we get,

$4=a \cos A$ and $-3=-a \sin A$

i.e. $a \cos A=4$ and $a \sin A=3$

i. e. $\frac{a \sin A}{a \cos A}=\frac{3}{4}$

i.e. $\tan A=\frac{3}{4}$

i.e. $A=\tan ^{-1}\left(\frac{3}{4}\right)$

and $(a \sin A)^{2}+(a \cos A)^{2}=9+16$

$a^{2}\left(\sin ^{2} A+\cos ^{2} A\right)=25$

i. e. $a^{2}=25$

i. e. $a=5$

$\therefore 4 \cos x-3 \sin x+7=5 \cos \left(x+\tan ^{-1}\left(\frac{3}{4}\right)+7\right)$

Since $\cos \theta=-1$ is minimum value of $\cos$

$\Rightarrow 4 \cos x-3 \sin x+7=5(-1)+7$

Minimum values for $4 \cos -3 \sin x+7=2$.