The equation $3 \cos x+4 \sin x=6$ has ..... solution.

Given that $3 \cos x+4 \sin x=6$
Since the maximum value of $(3 \cos x+4 \sin x)=+\sqrt{3^2+4^2}=5$. Hence there is no ' $x$ ' satisfying $3 \cos x+4 \sin x=6$. $\therefore$ No solution, as $|\sin x| \leq 1,|\cos x| \leq 1$ and both of them do not attain their maximum value for the same angle. Alternate Method, $3 \cos x+4 \sin x=6$ Let $r \cos \theta=3$ and $\quad r \sin \theta=4$ $\begin{array}{lr}\Rightarrow & \theta=\tan ^{-1}\left(\frac{4}{3}\right) \\ \Rightarrow & 5(\cos \theta \cos x+\sin \theta \sin x)=6\end{array}$ $\Rightarrow \cos (\theta-x)=\frac{6}{5}$ or $\cos (x-\theta)=\frac{6}{5}$ or $\cos (x-\theta)\gt1$ but $-1 \leq \cos x \leq 1$. $\therefore$ There exist no solution in $[-1,1]$.