If `4cos^(2)x^(@)-1=0 and 0 le x^(@) le 90^(@)`, find :
`x^(@)`

Answer

To solve the equation \(4 \cos^2 x - 1 = 0\) for \(0 \leq x \leq 90^\circ\), follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 4 \cos^2 x - 1 = 0 \]
Add 1 to both sides:
\[ 4 \cos^2 x = 1 \]

### Step 2: Dividing by 4
Now, divide both sides by 4:
\[ \cos^2 x = \frac{1}{4} \]

### Step 3: Taking the Square Root
Next, take the square root of both sides:
\[ \cos x = \pm \sqrt{\frac{1}{4}} \]
This simplifies to:
\[ \cos x = \pm \frac{1}{2} \]

### Step 4: Considering the Range of x
Since we are looking for solutions in the range \(0 \leq x \leq 90^\circ\), we only consider the positive value:
\[ \cos x = \frac{1}{2} \]

### Step 5: Finding the Angle
Now, we need to find the angle \(x\) for which \(\cos x = \frac{1}{2}\). From trigonometric ratios, we know:
\[ \cos 60^\circ = \frac{1}{2} \]
Thus,
\[ x = 60^\circ \]

### Final Answer
The value of \(x\) is:
\[ \boxed{60^\circ} \]

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