If 2 ` cos ^(2) theta + sin theta - 2=0 and 0^(@) le theta le 90 ^(@) ` find the value of ` theta. `

Answer

To solve the equation \(2 \cos^2 \theta + \sin \theta - 2 = 0\) for \(0^\circ \leq \theta \leq 90^\circ\), we will follow these steps:

### Step 1: Rewrite \(\cos^2 \theta\)
We know from the Pythagorean identity that:
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Substituting this into the equation gives:
\[ 2(1 - \sin^2 \theta) + \sin \theta - 2 = 0 \]

### Step 2: Simplify the equation
Expanding the equation:
\[ 2 - 2 \sin^2 \theta + \sin \theta - 2 = 0 \]
This simplifies to:
\[ -2 \sin^2 \theta + \sin \theta = 0 \]

### Step 3: Factor the equation
We can factor out \(\sin \theta\):
\[ \sin \theta (-2 \sin \theta + 1) = 0 \]

### Step 4: Set each factor to zero
This gives us two equations to solve:
1. \(\sin \theta = 0\)
2. \(-2 \sin \theta + 1 = 0\)

### Step 5: Solve \(\sin \theta = 0\)
For \(\sin \theta = 0\):
\[ \theta = 0^\circ \]

### Step 6: Solve \(-2 \sin \theta + 1 = 0\)
Rearranging gives:
\[ 2 \sin \theta = 1 \implies \sin \theta = \frac{1}{2} \]
The angle \(\theta\) that satisfies this in the range \(0^\circ \leq \theta \leq 90^\circ\) is:
\[ \theta = 30^\circ \]

### Step 7: Conclusion
Thus, the values of \(\theta\) that satisfy the original equation are:
\[ \theta = 0^\circ \quad \text{or} \quad \theta = 30^\circ \]