Question

The most general solution of $\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ is

• A. $\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6}$
• B. $\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{6}$
• C. $\theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6}$
• D. $\theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is C

$\theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6}$

Find the general solution of the given equation

In the question, an equation $\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ is given.

Divide both sides of the equation by $2$,

$$\begin{gathered} \frac{\sqrt{3}}{2}\cos \left(\theta \right)+\frac{1}{2}\sin \left(\theta \right)=\frac{\sqrt{2}}{2} \\ \Rightarrow \cos \left(\frac{\mathrm{\pi }}{6}\right)\cos \left(\theta \right)+\sin \left(\frac{\mathrm{\pi }}{6}\right)\sin \left(\theta \right)=\frac{1}{\sqrt{2}} \end{gathered}$$

We know that, $\cos (A-B)=\cos \left(A\right)\cos \left(B\right)+\sin \left(A\right)\sin \left(B\right)$ and

So,

$\cos \left(\theta -\frac{\mathrm{\pi }}{6}\right)=\cos \left(\frac{\mathrm{\pi }}{4}\right)$

This implies that,

$$\begin{gathered} \theta -\frac{\mathrm{\pi }}{6}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{\theta }=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6} \end{gathered}$$

Therefore, the general solution of the given equation is $\theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6},n\in Z$

Hence, option (C) is the correct answer.