Step 1: Understanding the Concept:
Two lines are parallel if they have the same slope. We need to find the point on the curve where the derivative (slope of the tangent) equals the slope of the given line. Then, formulate the equation of the tangent at that point.
Step 2: Key Formula or Approach:
1. Find slope $m$ of the line $x + 2y = 0$.
2. Differentiate $y = \cos(x+y)$ implicitly to find $\frac{dy}{dx}$.
3. Set $\frac{dy}{dx} = m$ to find the point(s) of tangency $(x_1, y_1)$.
4. Use point-slope form $y - y_1 = m(x - x_1)$ to write the equation.
Step 3: Detailed Explanation:
The given line is $x + 2y = 0$, which can be written as $y = -\frac{1}{2}x$. The slope $m = -\frac{1}{2}$.
Now, differentiate the curve $y = \cos(x + y)$ with respect to $x$:
\[ \frac{dy}{dx} = -\sin(x + y) \cdot \frac{d}{dx}(x + y) \]
\[ y' = -\sin(x + y) \cdot (1 + y') \]
Since the tangent is parallel to the line, substitute $y' = -\frac{1}{2}$:
\[ -\frac{1}{2} = -\sin(x + y) \cdot \left(1 - \frac{1}{2}\right) \]
\[ -\frac{1}{2} = -\sin(x + y) \cdot \left(\frac{1}{2}\right) \]
Divide both sides by $-1/2$:
\[ \sin(x + y) = 1 \]
This implies $x + y = \frac{\pi}{2} + 2n\pi$, for integer $n$.
Now substitute this back into the original curve equation to find the $y$-coordinate of the point of tangency:
\[ y = \cos(x + y) = \cos\left(\frac{\pi}{2} + 2n\pi\right) = 0 \]
Since $y = 0$, we can find $x$:
\[ x + 0 = \frac{\pi}{2} + 2n\pi \implies x = \frac{\pi}{2} + 2n\pi \]
We are given the domain $-2\pi \le x \le 2\pi$. Let's test values of $n$:
For $n = 0$: $x = \frac{\pi}{2}$. Point is $(\frac{\pi}{2}, 0)$.
For $n = -1$: $x = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2}$. Point is $(-\frac{3\pi}{2}, 0)$.
Other integer values of $n$ fall outside the domain.
Case 1: Tangent at point $(\frac{\pi}{2}, 0)$ with slope $-\frac{1}{2}$.
\[ y - 0 = -\frac{1}{2} \left(x - \frac{\pi}{2}\right) \]
Multiply by 4 to clear fractions and match option format:
\[ 4y = -2x + \pi \]
\[ 2x + 4y - \pi = 0 \]
This matches option (B).
Case 2: Tangent at point $(-\frac{3\pi}{2}, 0)$ with slope $-\frac{1}{2}$.
\[ y - 0 = -\frac{1}{2} \left(x - \left(-\frac{3\pi}{2}\right)\right) \]
\[ 2y = -x - \frac{3\pi}{2} \]
\[ 4y = -2x - 3\pi \implies 2x + 4y + 3\pi = 0 \]
This is not listed in the options.
Step 4: Final Answer:
The equation of the tangent is $2x + 4y - \pi = 0$.