If `cos 2x = cos 60^(@) cos 30^(@) + sin 60^(@) sin 30^(@)`, then the value of x is

Options

A. `15^(@)`
B. `40^(@)`
C. `30^(@)`
D. None of these

Answer

To solve the equation \( \cos 2x = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ \), we will follow these steps:

### Step 1: Use the Cosine Addition Formula
We can recognize that the right-hand side of the equation can be simplified using the cosine addition formula:
\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
In our case, we have:
\[ \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ = \cos(60^\circ - 30^\circ) = \cos 30^\circ \]

### Step 2: Substitute the Simplified Expression
Now we can substitute this back into our equation:
\[ \cos 2x = \cos 30^\circ \]

### Step 3: Find the Value of \( \cos 30^\circ \)
We know that:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Thus, we have:
\[ \cos 2x = \frac{\sqrt{3}}{2} \]

### Step 4: Solve for \( 2x \)
The cosine function equals \( \frac{\sqrt{3}}{2} \) at specific angles. The general solutions for \( \cos \theta = \frac{\sqrt{3}}{2} \) are:
\[ \theta = 30^\circ + 360^\circ n \quad \text{or} \quad \theta = 330^\circ + 360^\circ n \quad (n \in \mathbb{Z}) \]
So we have:
\[ 2x = 30^\circ + 360^\circ n \quad \text{or} \quad 2x = 330^\circ + 360^\circ n \]

### Step 5: Solve for \( x \)
Dividing both sides by 2 gives us:
\[ x = 15^\circ + 180^\circ n \quad \text{or} \quad x = 165^\circ + 180^\circ n \]

### Step 6: Find the Principal Value
For the principal value, we can take \( n = 0 \):
\[ x = 15^\circ \quad \text{or} \quad x = 165^\circ \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{15^\circ} \]