If `2cos^(2)x- sin^(2)x= -0.25 and 0^(@) le x lt 90^(@)`, then x= ?

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `45^(@)`

Answer

To solve the equation \(2\cos^2 x - \sin^2 x = -0.25\) for \(0^\circ \leq x < 90^\circ\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2\cos^2 x - \sin^2 x = -0.25 \]
We know that \(\cos^2 x = 1 - \sin^2 x\). We can substitute this into the equation:
\[ 2(1 - \sin^2 x) - \sin^2 x = -0.25 \]

### Step 2: Simplify the equation
Now, we simplify the equation:
\[ 2 - 2\sin^2 x - \sin^2 x = -0.25 \]
Combine the terms involving \(\sin^2 x\):
\[ 2 - 3\sin^2 x = -0.25 \]

### Step 3: Isolate \(\sin^2 x\)
Next, we isolate \(\sin^2 x\):
\[ -3\sin^2 x = -0.25 - 2 \]
\[ -3\sin^2 x = -2.25 \]
Dividing both sides by -3 gives:
\[ \sin^2 x = \frac{2.25}{3} = 0.75 \]

### Step 4: Solve for \(\sin x\)
Now we take the square root of both sides:
\[ \sin x = \pm \sqrt{0.75} = \pm \frac{\sqrt{3}}{2} \]
Since \(0^\circ \leq x < 90^\circ\), we only consider the positive value:
\[ \sin x = \frac{\sqrt{3}}{2} \]

### Step 5: Find the angle \(x\)
The angle \(x\) that satisfies \(\sin x = \frac{\sqrt{3}}{2}\) in the range \(0^\circ \leq x < 90^\circ\) is:
\[ x = 60^\circ \]

Thus, the solution is:
\[ \boxed{60^\circ} \]