Question

$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{real}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}3\sin \theta +4\cos \theta =6\mathrm{is}$
.

• A. 0.0
• B. 2
• C. 1
• D. Infinite

Answer 1

The correct option is A 0.0

Here, the equation is of the form $a\cos \theta +b\sin \theta =c$.
Now, dividing the equation by $\sqrt{a^2+b^2}=\sqrt{3^2+4^2}=5$ on both sides, we get the equation as:
$\frac{3}{5}\sin \theta +\frac{4}{5}\cos \theta =\frac{6}{5}$.
Now, assuming
$\cos \alpha =\frac{4}{5}\Rightarrow \sin \alpha =\frac{3}{5}$.
Now, we can write the equation as:
$\cos \theta \cos \theta +\sin \theta \sin \alpha =\frac{6}{5}\Rightarrow \cos (\theta -\alpha )=\frac{6}{5}>1.$
which is not possible as $cos\theta \in \left(-1,1\right].$Thus, the equation has no real roots.