In a triangle `ABC, cos 3A + cos 3B + cos 3C = 1`, then find any one angle.

Options

A. `pi/3`
B. `(2pi)/(3)`
C. `pi`
D. `(4pi)/(3)`

Answer

To solve the problem where \( \cos 3A + \cos 3B + \cos 3C = 1 \) in triangle \( ABC \), we can follow these steps:

### Step 1: Use the Angle Sum Property
In a triangle, the sum of the angles is \( A + B + C = 180^\circ \). We can express \( C \) in terms of \( A \) and \( B \):
\[ C = 180^\circ - A - B \]

### Step 2: Substitute into the Given Equation
Substituting \( C \) into the equation gives:
\[ \cos 3A + \cos 3B + \cos(3(180^\circ - A - B)) = 1 \]
Using the cosine identity \( \cos(180^\circ - x) = -\cos x \), we can rewrite \( \cos(3C) \):
\[ \cos(3C) = \cos(540^\circ - 3A - 3B) = -\cos(3A + 3B) \]
Thus, the equation becomes:
\[ \cos 3A + \cos 3B - \cos(3A + 3B) = 1 \]

### Step 3: Use the Cosine Addition Formula
Using the cosine addition formula \( \cos(A + B) = \cos A \cos B - \sin A \sin B \):
\[ \cos 3A + \cos 3B - (\cos 3A \cos 3B - \sin 3A \sin 3B) = 1 \]
This simplifies to:
\[ \cos 3A + \cos 3B - \cos 3A \cos 3B + \sin 3A \sin 3B = 1 \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ \cos 3A + \cos 3B + \sin 3A \sin 3B - \cos 3A \cos 3B = 1 \]
We can group terms:
\[ 1 - \cos 3A \cos 3B + \sin 3A \sin 3B = 1 \]
This implies:
\[ -\cos 3A \cos 3B + \sin 3A \sin 3B = 0 \]

### Step 5: Factor the Equation
Factoring gives:
\[ \sin 3A \sin 3B = \cos 3A \cos 3B \]
This can be rewritten as:
\[ \tan 3A = \tan 3B \]
Thus, we have:
\[ 3A = 3B + n\pi \quad \text{for some integer } n \]
This implies:
\[ A = B + \frac{n\pi}{3} \]

### Step 6: Solve for Angles
Since \( A + B + C = \pi \) (in radians), substituting \( A = B + \frac{n\pi}{3} \):
\[ (B + \frac{n\pi}{3}) + B + C = \pi \]
This simplifies to:
\[ 2B + C + \frac{n\pi}{3} = \pi \]
From here, we can find \( C \):
\[ C = \pi - 2B - \frac{n\pi}{3} \]

### Step 7: Find Specific Angles
Assuming \( n = 0 \):
\[ C = \pi - 2B \]
If we set \( B = \frac{\pi}{3} \):
\[ C = \pi - 2 \cdot \frac{\pi}{3} = \frac{\pi}{3} \]
Thus, \( A = B = \frac{\pi}{3} \) and \( C = \frac{\pi}{3} \).

If we set \( B = \frac{2\pi}{3} \):
\[ C = \pi - 2 \cdot \frac{2\pi}{3} = -\frac{\pi}{3} \quad \text{(not valid)} \]

### Conclusion
Thus, one angle \( C \) can be \( \frac{2\pi}{3} \).