The number of all possible 5-tuples `(a_(1),a_(2),a_(3),a_(4),a_(5))` such that ` a_(1)+a_(2) sin x+a_(3) cos x + a_(4) sin 2x +a_(5) cos 2 x =0` hold for all x is

Options

A. zero
B. 1
C. 2
D. infinite

Answer

To solve the problem of finding the number of all possible 5-tuples \((a_1, a_2, a_3, a_4, a_5)\) such that

\[ a_1 + a_2 \sin x + a_3 \cos x + a_4 \sin 2x + a_5 \cos 2x = 0 \]

holds for all \(x\), we can follow these steps:

### Step 1: Set up the equation for specific values of \(x\)

Since the equation must hold for all \(x\), we can evaluate it at specific values of \(x\) to derive relationships among the coefficients \(a_1, a_2, a_3, a_4,\) and \(a_5\).

### Step 2: Evaluate at \(x = 0\)

Substituting \(x = 0\):

\[ a_1 + a_2 \sin(0) + a_3 \cos(0) + a_4 \sin(0) + a_5 \cos(0) = 0 \]

This simplifies to:

\[ a_1 + a_3 + a_5 = 0 \quad \text{(Equation 1)} \]

### Step 3: Evaluate at \(x = \pi\)

Substituting \(x = \pi\):

\[ a_1 + a_2 \sin(\pi) + a_3 \cos(\pi) + a_4 \sin(2\pi) + a_5 \cos(2\pi) = 0 \]

This simplifies to:

\[ a_1 - a_3 + a_5 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve Equations 1 and 2

Now we have two equations:

1. \(a_1 + a_3 + a_5 = 0\)
2. \(a_1 - a_3 + a_5 = 0\)

Subtracting Equation 1 from Equation 2:

\[ (a_1 - a_3 + a_5) - (a_1 + a_3 + a_5) = 0 \]

This simplifies to:

\[ -2a_3 = 0 \implies a_3 = 0 \]

### Step 5: Substitute \(a_3 = 0\) back into the equations

Substituting \(a_3 = 0\) into Equation 1:

\[ a_1 + 0 + a_5 = 0 \implies a_1 + a_5 = 0 \quad \text{(Equation 3)} \]

### Step 6: Evaluate at \(x = \frac{\pi}{2}\)

Substituting \(x = \frac{\pi}{2}\):

\[ a_1 + a_2 \sin\left(\frac{\pi}{2}\right) + a_3 \cos\left(\frac{\pi}{2}\right) + a_4 \sin(\pi) + a_5 \cos(\pi) = 0 \]

This simplifies to:

\[ a_1 + a_2 + 0 + 0 - a_5 = 0 \implies a_1 + a_2 - a_5 = 0 \quad \text{(Equation 4)} \]

### Step 7: Substitute \(a_5\) from Equation 3 into Equation 4

From Equation 3, we have \(a_5 = -a_1\). Substituting this into Equation 4:

\[ a_1 + a_2 - (-a_1) = 0 \implies 2a_1 + a_2 = 0 \implies a_2 = -2a_1 \quad \text{(Equation 5)} \]

### Step 8: Evaluate at \(x = \frac{3\pi}{2}\)

Substituting \(x = \frac{3\pi}{2}\):

\[ a_1 + a_2 \sin\left(\frac{3\pi}{2}\right) + a_3 \cos\left(\frac{3\pi}{2}\right) + a_4 \sin(3\pi) + a_5 \cos(3\pi) = 0 \]

This simplifies to:

\[ a_1 - a_2 + 0 + 0 + a_5 = 0 \implies a_1 - a_2 + a_5 = 0 \quad \text{(Equation 6)} \]

### Step 9: Substitute \(a_5\) from Equation 3 into Equation 6

Substituting \(a_5 = -a_1\) into Equation 6:

\[ a_1 - a_2 - a_1 = 0 \implies -a_2 = 0 \implies a_2 = 0 \]

### Step 10: Substitute \(a_2 = 0\) back into Equation 5

From Equation 5, if \(a_2 = 0\):

\[ 0 = -2a_1 \implies a_1 = 0 \]

### Step 11: Substitute \(a_1 = 0\) into Equation 3

From Equation 3:

\[ 0 + a_5 = 0 \implies a_5 = 0 \]

### Step 12: Conclusion

Since \(a_3 = 0\), \(a_2 = 0\), \(a_1 = 0\), and \(a_5 = 0\), we still need to find \(a_4\). Since there are no restrictions on \(a_4\), it can take any value.

Thus, the only solution is:

\[ (a_1, a_2, a_3, a_4, a_5) = (0, 0, 0, a_4, 0) \]

where \(a_4\) can be any real number. Therefore, the number of all possible 5-tuples is infinite, as \(a_4\) can take any value.

### Final Answer

The number of all possible 5-tuples \((a_1, a_2, a_3, a_4, a_5)\) is infinite.

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