Question

$\sqrt{3}cos\theta + sin\theta = \sqrt{2}$

• A. $\theta = \frac{\pi}{2} + \frac{\pi}{6}$
• B. $\frac{\pi}{2} - \frac{\pi}{6}$
• C. $\frac{\pi}{5} - \frac{\pi}{3}$
• D. $\frac{\pi}{4} + \frac{\pi}{3}$

Answer 1

None of the given options are correct. The general solutions are

$$\theta=\frac{5\pi}{12}+2\pi k\quad \text{and}\quad \theta=-\frac{\pi}{12}+2\pi k,\quad k\in\mathbb{Z}.$$

Answer 2

Solution:

We start with the given equation:

$$\sqrt{3}\cos\theta+\sin\theta=\sqrt{2}.$$

Step 1. Write in the form $R\cos(\theta-\phi)$:

For an expression $a\cos\theta+b\sin\theta$ we have

$$R=\sqrt{a^2+b^2}=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{3+1}=2.$$

And

$$\cos\phi=\frac{a}{R}=\frac{\sqrt{3}}{2},\quad \sin\phi=\frac{1}{2}.$$

Thus, $\phi=\frac{\pi}{6}$.

So the equation becomes:

$$2\cos\Big(\theta-\frac{\pi}{6}\Big)=\sqrt{2}.$$

Step 2. Solve for $\theta$:

Divide both sides by 2:

$$\cos\Big(\theta-\frac{\pi}{6}\Big)=\frac{\sqrt{2}}{2}.$$

Recall that

$$\cos x=\frac{\sqrt{2}}{2}\quad\Longrightarrow\quad x=\frac{\pi}{4}+2\pi k\quad \text{or}\quad x=-\frac{\pi}{4}+2\pi k,\quad k\in\mathbb{Z}.$$

Thus,

$$\theta-\frac{\pi}{6}=\frac{\pi}{4}+2\pi k\quad\text{or}\quad \theta-\frac{\pi}{6}=-\frac{\pi}{4}+2\pi k.$$

So the solutions are:

$$\theta=\frac{\pi}{4}+\frac{\pi}{6}+2\pi k=\frac{5\pi}{12}+2\pi k,$$

and

$$\theta=-\frac{\pi}{4}+\frac{\pi}{6}+2\pi k=\frac{-3\pi+2\pi}{12}+2\pi k=\frac{-\pi}{12}+2\pi k.$$

Step 3. Compare with given options:

The provided options are:

1. $\theta=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3}$
2. $\theta=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$
3. $\theta=\frac{\pi}{5}-\frac{\pi}{3}$
4. $\theta=\frac{\pi}{4}+\frac{\pi}{3}=\frac{7\pi}{12}$

Neither $\frac{5\pi}{12}$ nor $-\frac{\pi}{12}$ matches any of these options.

Answer: None of the given options are correct. The general solutions are

$$\theta=\frac{5\pi}{12}+2\pi k\quad \text{and}\quad \theta=-\frac{\pi}{12}+2\pi k,\quad k\in\mathbb{Z}.$$