If `2cos theta - sin theta = 1/sqrt(2), (0^(@) lt theta lt 90^(@))` then the value of `2sin theta + cos theta` is:

Options

A. `1/sqrt(2)`
B. `sqrt(2)`
C. `3/sqrt(2)`
D. `sqrt(2)/3`

Answer

To solve the problem, we start with the equation given:

1. **Given Equation**:
\[ 2 \cos \theta - \sin \theta = \frac{1}{\sqrt{2}} \]

2. **Rearranging the Equation**:
We can rearrange this equation to express \(\sin \theta\) in terms of \(\cos \theta\):
\[ \sin \theta = 2 \cos \theta - \frac{1}{\sqrt{2}} \]

3. **Finding \(2 \sin \theta + \cos \theta\)**:
We need to find the value of \(2 \sin \theta + \cos \theta\). Substituting \(\sin \theta\) from the previous step:
\[ 2 \sin \theta + \cos \theta = 2(2 \cos \theta - \frac{1}{\sqrt{2}}) + \cos \theta \]
\[ = 4 \cos \theta - \frac{2}{\sqrt{2}} + \cos \theta \]
\[ = 5 \cos \theta - \sqrt{2} \]

4. **Using the Pythagorean Identity**:
We know that \(\sin^2 \theta + \cos^2 \theta = 1\). Substituting \(\sin \theta\) from step 2:
\[ (2 \cos \theta - \frac{1}{\sqrt{2}})^2 + \cos^2 \theta = 1 \]

5. **Expanding the Equation**:
Expanding the left-hand side:
\[ (4 \cos^2 \theta - 2 \cdot 2 \cos \theta \cdot \frac{1}{\sqrt{2}} + \frac{1}{2}) + \cos^2 \theta = 1 \]
\[ 4 \cos^2 \theta - \frac{4 \cos \theta}{\sqrt{2}} + \frac{1}{2} + \cos^2 \theta = 1 \]
\[ 5 \cos^2 \theta - \frac{4 \cos \theta}{\sqrt{2}} + \frac{1}{2} - 1 = 0 \]
\[ 5 \cos^2 \theta - \frac{4 \cos \theta}{\sqrt{2}} - \frac{1}{2} = 0 \]

6. **Solving the Quadratic Equation**:
Let \(x = \cos \theta\). The equation becomes:
\[ 5x^2 - \frac{4x}{\sqrt{2}} - \frac{1}{2} = 0 \]
Multiplying through by \(2\) to eliminate the fraction:
\[ 10x^2 - 4\sqrt{2}x - 1 = 0 \]

7. **Using the Quadratic Formula**:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{4\sqrt{2} \pm \sqrt{(4\sqrt{2})^2 - 4 \cdot 10 \cdot (-1)}}{2 \cdot 10} \]
\[ = \frac{4\sqrt{2} \pm \sqrt{32 + 40}}{20} \]
\[ = \frac{4\sqrt{2} \pm \sqrt{72}}{20} \]
\[ = \frac{4\sqrt{2} \pm 6\sqrt{2}}{20} \]
\[ = \frac{10\sqrt{2}}{20} \quad \text{or} \quad \frac{-2\sqrt{2}}{20} \]
\[ = \frac{\sqrt{2}}{2} \quad \text{(valid since } \theta \text{ is in the first quadrant)} \]

8. **Finding \(2 \sin \theta + \cos \theta\)**:
Now substituting \(\cos \theta = \frac{\sqrt{2}}{2}\) back into \(2 \sin \theta + \cos \theta\):
\[ 2 \sin \theta + \frac{\sqrt{2}}{2} = 5 \cdot \frac{\sqrt{2}}{2} - \sqrt{2} \]
\[ = \frac{5\sqrt{2}}{2} - \sqrt{2} = \frac{5\sqrt{2}}{2} - \frac{2\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \]

Thus, the final answer is:
\[ \boxed{\frac{3\sqrt{2}}{2}} \]