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 can follow these steps:

### Step 1: Substitute the values of trigonometric functions

We know the following values:
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)

Now, substituting these values into the equation:

\[ \cos 2x = \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right) \]

### Step 2: Simplify the right-hand side

Calculating the right-hand side:

\[ \cos 2x = \frac{1 \cdot \sqrt{3}}{4} + \frac{\sqrt{3} \cdot 1}{4} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]

### Step 3: Set the equation

Now we have:

\[ \cos 2x = \frac{\sqrt{3}}{2} \]

### Step 4: Find the angle corresponding to the cosine value

The cosine function equals \( \frac{\sqrt{3}}{2} \) at specific angles. The principal angle is:

\[ 2x = 30^\circ \quad \text{or} \quad 2x = 360^\circ - 30^\circ = 330^\circ \]

### Step 5: Solve for \( x \)

Dividing both angles by 2 to find \( x \):

1. From \( 2x = 30^\circ \):
\[ x = \frac{30^\circ}{2} = 15^\circ \]

2. From \( 2x = 330^\circ \):
\[ x = \frac{330^\circ}{2} = 165^\circ \]

### Step 6: Conclusion

The possible values of \( x \) are \( 15^\circ \) and \( 165^\circ \). However, since we are looking for a value of \( x \) that is typically within the range of \( 0^\circ \) to \( 180^\circ \), we can conclude that:

\[ \text{The value of } x \text{ is } 15^\circ. \]

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