If `sqrt(3)costheta+sintheta=2` , then general value of `theta` is:

Options

A. `(a)` `npi+(-1)^npi/2,\ ninZ`
B. `(b)`. `(-1)^npi/2-pi/3,\ ninZ`
C. `(c)`. `npi+-pi/2-pi/3, ninZ`
D. `(d)`. `npi+(-1)^npi/2-pi/3, ninZ`

Answer

To solve the equation \( \sqrt{3} \cos \theta + \sin \theta = 2 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sqrt{3} \cos \theta + \sin \theta = 2 \]
We can rearrange this equation to isolate the trigonometric terms:
\[ \sqrt{3} \cos \theta + \sin \theta - 2 = 0 \]

### Step 2: Normalize the coefficients
Next, we divide the entire equation by 2 to make it easier to work with:
\[ \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = 1 \]

### Step 3: Identify sine and cosine values
We recognize that \( \frac{\sqrt{3}}{2} \) is the sine of \( 60^\circ \) (or \( \frac{\pi}{3} \)) and \( \frac{1}{2} \) is the cosine of \( 60^\circ \). Thus, we can express the equation in terms of sine:
\[ \sin\left(\frac{\pi}{3}\right) \cos \theta + \cos\left(\frac{\pi}{3}\right) \sin \theta = 1 \]

### Step 4: Use the sine addition formula
Using the sine addition formula \( \sin(a + b) = \sin a \cos b + \cos a \sin b \), we can rewrite the equation:
\[ \sin\left(\frac{\pi}{3} + \theta\right) = 1 \]

### Step 5: Solve for the angle
The general solution for \( \sin x = 1 \) is given by:
\[ x = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
\[ \frac{\pi}{3} + \theta = \frac{\pi}{2} + 2n\pi \]

### Step 6: Isolate \( \theta \)
Now, we isolate \( \theta \):
\[ \theta = \frac{\pi}{2} - \frac{\pi}{3} + 2n\pi \]
Calculating \( \frac{\pi}{2} - \frac{\pi}{3} \):
\[ \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{6} \]
Thus, we have:
\[ \theta = \frac{\pi}{6} + 2n\pi \]

### Step 7: Generalize the solution
Since the sine function is periodic, we also consider the negative sine solution:
\[ \theta = n\pi + (-1)^n \frac{\pi}{2} - \frac{\pi}{3} \]

### Final Answer
The general value of \( \theta \) is:
\[ \theta = n\pi + (-1)^n \frac{\pi}{2} - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]