When `2sin^(2)theta=3costheta`, and `0lethetale90^(@)`, then`theta`=?

Options

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

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\sin^2\theta = 3\cos\theta \]
Using the identity \(\sin^2\theta = 1 - \cos^2\theta\), we can substitute for \(\sin^2\theta\):
\[ 2(1 - \cos^2\theta) = 3\cos\theta \]

### Step 2: Expand and rearrange
Expanding the left side gives us:
\[ 2 - 2\cos^2\theta = 3\cos\theta \]
Rearranging this equation leads to:
\[ 2\cos^2\theta + 3\cos\theta - 2 = 0 \]

### Step 3: Identify coefficients for the quadratic equation
This is a quadratic equation in the form \(Ax^2 + Bx + C = 0\), where:
- \(A = 2\)
- \(B = 3\)
- \(C = -2\)

### Step 4: Use the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting in our values:
\[ \cos\theta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]

### Step 5: Calculate the discriminant
Calculating the discriminant:
\[ B^2 - 4AC = 9 + 16 = 25 \]
Thus, we have:
\[ \cos\theta = \frac{-3 \pm \sqrt{25}}{4} \]

### Step 6: Solve for \(\cos\theta\)
Calculating the square root gives us:
\[ \cos\theta = \frac{-3 \pm 5}{4} \]
This results in two potential solutions:
1. \(\cos\theta = \frac{2}{4} = \frac{1}{2}\)
2. \(\cos\theta = \frac{-8}{4} = -2\) (not valid since cosine cannot be less than -1)

### Step 7: Find \(\theta\)
Since \(\cos\theta = \frac{1}{2}\), we find:
\[ \theta = 60^\circ \]

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