If `0^(@)lt x lt 90^(@)` and `5 sin^(2)x = 7 sin x - 2`, what is the vlaue of sin x ?

Options

A. `1.00`
B. `0.71`
C. `0.40`
D. `0.38`

Explanation

To solve the equation \(5 \sin^2 x = 7 \sin x - 2\) for \( \sin x \), we can follow these steps:

### Step 1: Substitute \(y\) for \(\sin x\)
Let \(y = \sin x\). The equation becomes:
\[ 5y^2 = 7y - 2 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ 5y^2 - 7y + 2 = 0 \]

### Step 3: Identify coefficients
In the quadratic equation \(ay^2 + by + c = 0\), we identify:
- \(a = 5\)
- \(b = -7\)
- \(c = 2\)

### Step 4: Apply the quadratic formula
The quadratic formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 5 \cdot 2}}{2 \cdot 5} \]

### Step 5: Simplify the expression
Calculating the discriminant:
\[ y = \frac{7 \pm \sqrt{49 - 40}}{10} \]
\[ y = \frac{7 \pm \sqrt{9}}{10} \]
\[ y = \frac{7 \pm 3}{10} \]

### Step 6: Calculate the possible values for \(y\)
Now we calculate the two possible values:
1. \(y = \frac{7 + 3}{10} = \frac{10}{10} = 1\)
2. \(y = \frac{7 - 3}{10} = \frac{4}{10} = 0.4\)

### Step 7: Determine valid solutions
Since \(0 < \sin x < 1\) for \(0^\circ < x < 90^\circ\), we discard \(y = 1\) because it corresponds to \(x = 90^\circ\), which is not in the interval. Thus, the only valid solution is:
\[ \sin x = 0.4 \]

### Final Answer
The value of \(\sin x\) is \(0.4\).

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