The value of `theta(0lethetale90^(@))` satisfying `2sin^(2)theta=3costheta` is

Options

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

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\sin^2\theta = 3\cos\theta \]
We can use the Pythagorean identity \(\sin^2\theta = 1 - \cos^2\theta\) to express everything in terms of \(\cos\theta\).

### Step 2: Substitute the identity
Substituting \(\sin^2\theta\) gives us:
\[ 2(1 - \cos^2\theta) = 3\cos\theta \]

### Step 3: Expand the equation
Expanding the left-hand side:
\[ 2 - 2\cos^2\theta = 3\cos\theta \]

### Step 4: Rearrange the equation
Rearranging the equation to one side gives:
\[ 2\cos^2\theta + 3\cos\theta - 2 = 0 \]

### Step 5: Factor the quadratic equation
Now we can factor this quadratic equation. We look for two numbers that multiply to \(2 \times -2 = -4\) and add to \(3\). The numbers \(4\) and \(-1\) work:
\[ (2\cos\theta - 1)(\cos\theta + 2) = 0 \]

### Step 6: Solve for \(\cos\theta\)
Setting each factor to zero gives us:
1. \(2\cos\theta - 1 = 0 \Rightarrow \cos\theta = \frac{1}{2}\)
2. \(\cos\theta + 2 = 0 \Rightarrow \cos\theta = -2\) (not possible since \(\cos\theta\) must be between -1 and 1)

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

### Conclusion
Thus, the value of \(\theta\) that satisfies the equation \(2\sin^2\theta = 3\cos\theta\) is:
\[ \theta = 60^\circ \]

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