The equation 3 cos x+4 sin x=6 has

(d) no

Given equation: 

$3 \cos x+4 \sin x=6 \quad \ldots(\mathrm{i})$

Thus, the equation is of the form $a \cos x+b \sin x=c$, where $a=3, b=4$ and $c=6$.

Let:

$a=3=r \cos \alpha$ and $b=4=r \sin \alpha$

Now,

$\tan \alpha=\frac{b}{a}=\frac{4}{3}$

$\Rightarrow \alpha=\tan ^{-1}\left(\frac{4}{3}\right)$

Aso,

$r=\sqrt{a^{2}+b^{2}}=\sqrt{9+16}=\sqrt{25}=5$

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

$r \cos \alpha \cos \theta+\sin \alpha \sin \theta=6$

$\Rightarrow r \cos (\theta-\alpha)=6$

$\Rightarrow 5 \cos (\theta-\alpha)=6$

$\Rightarrow \cos (\theta-\alpha)=\frac{6}{5}$

From here, we cannot find the value of $\theta$.