If `3 cos^(2) A + 7 sin ^(2) A = 3, 0^(@) le A le 90^(@) ` then the value of A is `:`

Options

A. `90^(@)`
B. `45^(@)`
C. `0^(@)`
D. `30^(@)`

Answer

To solve the equation \(3 \cos^2 A + 7 \sin^2 A = 3\) for \(A\) in the range \(0^\circ \leq A \leq 90^\circ\), we can follow these steps:

### Step 1: Use the Pythagorean Identity
We know from trigonometric identities that:
\[ \sin^2 A + \cos^2 A = 1 \]
We can express \(\sin^2 A\) in terms of \(\cos^2 A\):
\[ \sin^2 A = 1 - \cos^2 A \]

### Step 2: Substitute into the Equation
Substituting \(\sin^2 A\) into the original equation gives:
\[ 3 \cos^2 A + 7 (1 - \cos^2 A) = 3 \]

### Step 3: Simplify the Equation
Expanding the equation:
\[ 3 \cos^2 A + 7 - 7 \cos^2 A = 3 \]
Combine like terms:
\[ (3 - 7) \cos^2 A + 7 = 3 \]
This simplifies to:
\[ -4 \cos^2 A + 7 = 3 \]

### Step 4: Isolate \(\cos^2 A\)
Rearranging gives:
\[ -4 \cos^2 A = 3 - 7 \]
\[ -4 \cos^2 A = -4 \]
Dividing both sides by -4:
\[ \cos^2 A = 1 \]

### Step 5: Solve for \(\cos A\)
Taking the square root of both sides:
\[ \cos A = 1 \]
Since \(A\) is in the range \(0^\circ \leq A \leq 90^\circ\), we find:
\[ A = 0^\circ \]

### Final Answer
Thus, the value of \(A\) is:
\[ \boxed{0^\circ} \]