If `cos18^(@)-sin18^(@)=sqrt(x)sin27^(@)`, then `x=`

Answer

To solve the equation \( \cos 18^\circ - \sin 18^\circ = \sqrt{x} \sin 27^\circ \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \cos 18^\circ - \sin 18^\circ = \sqrt{x} \sin 27^\circ \]

### Step 2: Use the identity for sine
Recall that \( \sin(90^\circ - \theta) = \cos \theta \). We can express \( \sin 18^\circ \) as:
\[ \sin 18^\circ = \cos(90^\circ - 18^\circ) = \cos 72^\circ \]
Thus, we can rewrite the left-hand side:
\[ \cos 18^\circ - \cos 72^\circ = \sqrt{x} \sin 27^\circ \]

### Step 3: Apply the cosine difference formula
We can use the cosine difference formula:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Let \( A = 18^\circ \) and \( B = 72^\circ \):
\[ \cos 18^\circ - \cos 72^\circ = -2 \sin\left(\frac{18^\circ + 72^\circ}{2}\right) \sin\left(\frac{18^\circ - 72^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{18^\circ + 72^\circ}{2} = 45^\circ \quad \text{and} \quad \frac{18^\circ - 72^\circ}{2} = -27^\circ \]
Thus, we have:
\[ \cos 18^\circ - \cos 72^\circ = -2 \sin(45^\circ) \sin(-27^\circ) \]

### Step 4: Simplify the sine terms
Using the identity \( \sin(-\theta) = -\sin(\theta) \):
\[ \sin(-27^\circ) = -\sin(27^\circ) \]
So,
\[ \cos 18^\circ - \cos 72^\circ = -2 \sin(45^\circ)(-\sin(27^\circ)) = 2 \sin(45^\circ) \sin(27^\circ) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \cos 18^\circ - \cos 72^\circ = 2 \cdot \frac{1}{\sqrt{2}} \sin(27^\circ) = \sqrt{2} \sin(27^\circ) \]

### Step 5: Set the equations equal
Now we can equate both sides:
\[ \sqrt{2} \sin(27^\circ) = \sqrt{x} \sin(27^\circ) \]

### Step 6: Cancel \(\sin(27^\circ)\)
Assuming \( \sin(27^\circ) \neq 0 \), we can divide both sides by \( \sin(27^\circ) \):
\[ \sqrt{2} = \sqrt{x} \]

### Step 7: Solve for \(x\)
Squaring both sides gives:
\[ 2 = x \]

Thus, the final answer is:
\[ \boxed{2} \]