Write the number of solutions of the equation

We have :

$4 \sin x-3 \cos x=7 \ldots$..(i)

The equation is of the form $a \sin x+b \cos x=c$, where $a=4, b=-3$ and $c=7$.
Now,

Let:

$a=r \sin \alpha$ and $b=r \cos \alpha$

Thus, we have:

$r=\sqrt{a^{2}+b^{2}}=\sqrt{4^{2}+3^{2}}=5$ and $\tan \alpha=\frac{-4}{3} \Rightarrow \alpha=\tan ^{-1}\left(-\frac{4}{3}\right)$

By putting $a=4=r \sin \alpha$ and $b=-3=r \cos \alpha$ in equation (i), we get:

$r \sin \alpha \sin x+r \cos \alpha \cos x=7$

$\Rightarrow r \cos (x-\alpha)=7$

$\Rightarrow 5 \cos \left[x-\tan ^{-1}\left(\frac{-4}{3}\right)\right]=7$

$\Rightarrow \cos \left[\mathrm{x}-\tan ^{-1}\left(\frac{-4}{3}\right)\right]=\frac{7}{5}$

The solution is not possible.

Hence, the given equation has no solution.