If `0^@ lt theta lt 90^@` , find `theta` when 2 cos `theta+2sqrt2=3sec theta`

Answer

To solve the equation \( 2 \cos \theta + 2\sqrt{2} = 3 \sec \theta \) for \( \theta \) in the range \( 0^\circ < \theta < 90^\circ \), follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \sec \theta = \frac{1}{\cos \theta} \). Thus, we can rewrite the equation as:
\[ 2 \cos \theta + 2\sqrt{2} = \frac{3}{\cos \theta} \]

### Step 2: Multiply both sides by \( \cos \theta \)
To eliminate the fraction, multiply both sides by \( \cos \theta \):
\[ 2 \cos^2 \theta + 2\sqrt{2} \cos \theta = 3 \]

### Step 3: Rearrange the equation
Rearranging gives us a standard quadratic form:
\[ 2 \cos^2 \theta + 2\sqrt{2} \cos \theta - 3 = 0 \]

### Step 4: Identify coefficients for the quadratic formula
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 2 \)
- \( b = 2\sqrt{2} \)
- \( c = -3 \)

### Step 5: Apply the quadratic formula
The quadratic formula is given by:
\[ \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting in our values:
\[ \cos \theta = \frac{-2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \]

### Step 6: Simplify the expression under the square root
Calculating \( b^2 - 4ac \):
\[ (2\sqrt{2})^2 = 8 \]
\[ -4 \cdot 2 \cdot (-3) = 24 \]
Thus,
\[ b^2 - 4ac = 8 + 24 = 32 \]

### Step 7: Substitute back into the formula
Now substituting back:
\[ \cos \theta = \frac{-2\sqrt{2} \pm \sqrt{32}}{4} \]
Since \( \sqrt{32} = 4\sqrt{2} \), we have:
\[ \cos \theta = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{4} \]

### Step 8: Solve for the two potential values of \( \cos \theta \)
This gives us two potential solutions:
1. \( \cos \theta = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} \)
2. \( \cos \theta = \frac{-6\sqrt{2}}{4} = -\frac{3\sqrt{2}}{2} \) (not valid since cosine cannot be negative in the first quadrant)

### Step 9: Find the angle \( \theta \)
From \( \cos \theta = \frac{\sqrt{2}}{2} \), we find:
\[ \theta = 45^\circ \]

### Conclusion
Thus, the value of \( \theta \) is:
\[ \theta = 45^\circ \]