If `0^(@)lexle90^(@)` and `sinx+sqrt(3)cosx=1` , then what is the value of `x` ?

Options

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

Answer

To solve the equation \( \sin x + \sqrt{3} \cos x = 1 \) for \( x \) in the range \( 0^\circ \leq x \leq 90^\circ \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin x + \sqrt{3} \cos x = 1 \]

### Step 2: Isolate the sine term
To isolate \( \sin x \), we can rearrange the equation:
\[ \sin x = 1 - \sqrt{3} \cos x \]

### Step 3: Square both sides
To eliminate the sine and cosine, we can square both sides:
\[ (\sin x)^2 = (1 - \sqrt{3} \cos x)^2 \]
This expands to:
\[ \sin^2 x = 1 - 2\sqrt{3} \cos x + 3 \cos^2 x \]

### Step 4: Use the Pythagorean identity
Recall that \( \sin^2 x + \cos^2 x = 1 \), so we can substitute \( \sin^2 x \) with \( 1 - \cos^2 x \):
\[ 1 - \cos^2 x = 1 - 2\sqrt{3} \cos x + 3 \cos^2 x \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 0 = 4 \cos^2 x - 2\sqrt{3} \cos x \]

### Step 6: Factor the equation
Factoring out \( \cos x \):
\[ \cos x (4 \cos x - 2\sqrt{3}) = 0 \]

### Step 7: Solve for \( \cos x \)
This gives us two cases:
1. \( \cos x = 0 \)
2. \( 4 \cos x - 2\sqrt{3} = 0 \)

From the first case, \( \cos x = 0 \) leads to:
\[ x = 90^\circ \]

From the second case:
\[ 4 \cos x = 2\sqrt{3} \implies \cos x = \frac{\sqrt{3}}{2} \]
This leads to:
\[ x = 30^\circ \]

### Step 8: Check the solutions
Now we need to check which of these solutions satisfies the original equation:
- For \( x = 30^\circ \):
\[ \sin 30^\circ + \sqrt{3} \cos 30^\circ = \frac{1}{2} + \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} + \frac{3}{2} = 2 \quad (\text{not } 1) \]
- For \( x = 90^\circ \):
\[ \sin 90^\circ + \sqrt{3} \cos 90^\circ = 1 + 0 = 1 \quad (\text{correct}) \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{90^\circ} \]