For what value of `alpha` does the equation `4cosalpha+3cos2alpha-2sin3alpha+cos4alpha=2sqrt(3)-1` hold?

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(pi)/(2)`

Answer

To solve the equation \( 4\cos\alpha + 3\cos2\alpha - 2\sin3\alpha + \cos4\alpha = 2\sqrt{3} - 1 \), we will evaluate the possible values of \( \alpha \) given in the options: \( \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{2} \).

### Step 1: Evaluate \( \alpha = \frac{\pi}{6} \)

1. **Calculate \( \cos\frac{\pi}{6} \)**:
\[ \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \]

2. **Calculate \( \cos2\frac{\pi}{6} = \cos\frac{\pi}{3} \)**:
\[ \cos\frac{\pi}{3} = \frac{1}{2} \]

3. **Calculate \( \sin3\frac{\pi}{6} = \sin\frac{\pi}{2} \)**:
\[ \sin\frac{\pi}{2} = 1 \]

4. **Calculate \( \cos4\frac{\pi}{6} = \cos\frac{2\pi}{3} \)**:
\[ \cos\frac{2\pi}{3} = -\frac{1}{2} \]

5. **Substitute these values into the equation**:
\[ 4\left(\frac{\sqrt{3}}{2}\right) + 3\left(\frac{1}{2}\right) - 2(1) + \left(-\frac{1}{2}\right) \]

6. **Simplify the expression**:
\[ = 2\sqrt{3} + \frac{3}{2} - 2 - \frac{1}{2} \]
\[ = 2\sqrt{3} + \frac{3}{2} - \frac{4}{2} = 2\sqrt{3} - \frac{1}{2} \]

### Step 2: Compare with the right-hand side

We need to check if:
\[ 2\sqrt{3} - \frac{1}{2} = 2\sqrt{3} - 1 \]

This simplifies to:
\[ -\frac{1}{2} \neq -1 \]
So, \( \alpha = \frac{\pi}{6} \) does not satisfy the equation.

### Step 3: Evaluate \( \alpha = \frac{\pi}{3} \)

1. **Calculate \( \cos\frac{\pi}{3} \)**:
\[ \cos\frac{\pi}{3} = \frac{1}{2} \]

2. **Calculate \( \cos2\frac{\pi}{3} = \cos\frac{2\pi}{3} \)**:
\[ \cos\frac{2\pi}{3} = -\frac{1}{2} \]

3. **Calculate \( \sin3\frac{\pi}{3} = \sin\pi \)**:
\[ \sin\pi = 0 \]

4. **Calculate \( \cos4\frac{\pi}{3} = \cos\frac{4\pi}{3} \)**:
\[ \cos\frac{4\pi}{3} = -\frac{1}{2} \]

5. **Substitute these values into the equation**:
\[ 4\left(\frac{1}{2}\right) + 3\left(-\frac{1}{2}\right) - 2(0) + \left(-\frac{1}{2}\right) \]

6. **Simplify the expression**:
\[ = 2 - \frac{3}{2} - 0 - \frac{1}{2} \]
\[ = 2 - 2 = 0 \]

### Step 4: Compare with the right-hand side

We need to check if:
\[ 0 = 2\sqrt{3} - 1 \]
This does not hold true, so \( \alpha = \frac{\pi}{3} \) does not satisfy the equation.

### Step 5: Evaluate \( \alpha = \frac{\pi}{4} \)

1. **Calculate \( \cos\frac{\pi}{4} \)**:
\[ \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \]

2. **Calculate \( \cos2\frac{\pi}{4} = \cos\frac{\pi}{2} \)**:
\[ \cos\frac{\pi}{2} = 0 \]

3. **Calculate \( \sin3\frac{\pi}{4} \)**:
\[ \sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}} \]

4. **Calculate \( \cos4\frac{\pi}{4} = \cos\pi \)**:
\[ \cos\pi = -1 \]

5. **Substitute these values into the equation**:
\[ 4\left(\frac{1}{\sqrt{2}}\right) + 3(0) - 2\left(\frac{1}{\sqrt{2}}\right) + (-1) \]

6. **Simplify the expression**:
\[ = \frac{4}{\sqrt{2}} - \frac{2}{\sqrt{2}} - 1 \]
\[ = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \]

### Step 6: Compare with the right-hand side

We need to check if:
\[ \sqrt{2} - 1 = 2\sqrt{3} - 1 \]
This does not hold true, so \( \alpha = \frac{\pi}{4} \) does not satisfy the equation.

### Step 7: Evaluate \( \alpha = \frac{\pi}{2} \)

1. **Calculate \( \cos\frac{\pi}{2} \)**:
\[ \cos\frac{\pi}{2} = 0 \]

2. **Calculate \( \cos2\frac{\pi}{2} = \cos\pi \)**:
\[ \cos\pi = -1 \]

3. **Calculate \( \sin3\frac{\pi}{2} \)**:
\[ \sin\frac{3\pi}{2} = -1 \]

4. **Calculate \( \cos4\frac{\pi}{2} = \cos2\pi \)**:
\[ \cos2\pi = 1 \]

5. **Substitute these values into the equation**:
\[ 4(0) + 3(-1) - 2(-1) + 1 \]

6. **Simplify the expression**:
\[ = 0 - 3 + 2 + 1 = 0 \]

### Step 8: Compare with the right-hand side

We need to check if:
\[ 0 = 2\sqrt{3} - 1 \]
This does not hold true, so \( \alpha = \frac{\pi}{2} \) does not satisfy the equation.

### Conclusion

After evaluating all options, we find that the only value that satisfies the equation is:
\[ \alpha = \frac{\pi}{6} \]