If `0^(@)ltthetalt90^(@)and2sin^(2)theta+3costheta=3` , then the value of `theta` is

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `75^(@)`

Answer

To solve the equation \(2\sin^2\theta + 3\cos\theta = 3\) under the condition \(0^\circ < \theta < 90^\circ\), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 2\sin^2\theta + 3\cos\theta = 3 \]
We can rearrange it to isolate the trigonometric functions:
\[ 2\sin^2\theta = 3 - 3\cos\theta \]

### Step 2: Use the Pythagorean identity
Recall the Pythagorean identity:
\[ \sin^2\theta + \cos^2\theta = 1 \]
From this, we can express \(\sin^2\theta\) in terms of \(\cos\theta\):
\[ \sin^2\theta = 1 - \cos^2\theta \]
Substituting this into our rearranged equation gives:
\[ 2(1 - \cos^2\theta) = 3 - 3\cos\theta \]

### Step 3: Simplify the equation
Expanding and simplifying:
\[ 2 - 2\cos^2\theta = 3 - 3\cos\theta \]
Rearranging gives:
\[ 2\cos^2\theta - 3\cos\theta + 1 = 0 \]

### Step 4: Solve the quadratic equation
Now we have a quadratic equation in \(\cos\theta\):
\[ 2x^2 - 3x + 1 = 0 \]
where \(x = \cos\theta\). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 2\), \(b = -3\), and \(c = 1\):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ x = \frac{3 \pm \sqrt{9 - 8}}{4} \]
\[ x = \frac{3 \pm 1}{4} \]
This gives us two solutions:
\[ x = \frac{4}{4} = 1 \quad \text{and} \quad x = \frac{2}{4} = \frac{1}{2} \]

### Step 5: Determine the angles
Now we find \(\theta\) for these values of \(x\):
1. For \(x = 1\):
\(\cos\theta = 1\) implies \(\theta = 0^\circ\) (not in the range).
2. For \(x = \frac{1}{2}\):
\(\cos\theta = \frac{1}{2}\) implies \(\theta = 60^\circ\).

### Conclusion
Thus, the value of \(\theta\) is:
\[ \theta = 60^\circ \]