If `0ltxltpiand cosx+sinx=1/2,` then tan x is equal to

Options

A. `(1-sqrt7)/(4)`
B. `(4-sqrt7)/(3)`
C. `(-(4+sqrt7))/(3)`
D. `(sqrt7+1)/(4)`

Explanation

To solve the problem where \( \cos x + \sin x = \frac{1}{2} \) and we need to find \( \tan x \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ \cos x + \sin x = \frac{1}{2} \]

### Step 2: Divide the entire equation by \( \cos x \)
This gives us:
\[ 1 + \frac{\sin x}{\cos x} = \frac{1}{2 \cos x} \]
This can be rewritten as:
\[ 1 + \tan x = \frac{1}{2 \cos x} \]

### Step 3: Rewrite the equation in terms of \( \tan x \)
Let’s rearrange the equation:
\[ \tan x = \frac{1}{2 \cos x} - 1 \]

### Step 4: Square both sides
To eliminate the tangent, we square both sides:
\[ (1 + \tan x)^2 = \left(\frac{1}{2 \cos x}\right)^2 \]
Expanding the left side:
\[ 1 + 2\tan x + \tan^2 x = \frac{1}{4 \cos^2 x} \]

### Step 5: Use the identity \( \sec^2 x = 1 + \tan^2 x \)
We know that \( \sec^2 x = 1 + \tan^2 x \), so we can substitute:
\[ 1 + 2\tan x + \tan^2 x = \frac{1}{4 \cos^2 x} \]
This implies:
\[ 1 + 2\tan x + (1 + \tan^2 x) = \frac{1}{4 \cos^2 x} \]

### Step 6: Rearranging the equation
Now we can simplify:
\[ 2 + 2\tan x + \tan^2 x = \frac{1}{4 \cos^2 x} \]

### Step 7: Substitute \( \sec^2 x \)
Using \( \sec^2 x = 1 + \tan^2 x \):
\[ 2 + 2\tan x + \tan^2 x = \frac{1 + \tan^2 x}{4} \]

### Step 8: Clear the fraction by multiplying through by 4
This gives:
\[ 8 + 8\tan x + 4\tan^2 x = 1 + \tan^2 x \]

### Step 9: Rearranging to form a quadratic equation
Rearranging gives:
\[ 3\tan^2 x + 8\tan x + 7 = 0 \]

### Step 10: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \( \tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 8, c = 7 \):
\[ \tan x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3} \]
Calculating the discriminant:
\[ \tan x = \frac{-8 \pm \sqrt{64 - 84}}{6} \]
\[ \tan x = \frac{-8 \pm \sqrt{-20}}{6} \]

### Step 11: Simplifying the result
Since we have a negative discriminant, we can express it as:
\[ \tan x = \frac{-8 \pm 2i\sqrt{5}}{6} \]
\[ \tan x = \frac{-4 \pm i\sqrt{5}}{3} \]

### Final Result
Since \( x \) is in the range \( (0, \pi) \), we can conclude:
\[ \tan x = \frac{-4 + \sqrt{7}}{3} \quad \text{(valid solution)} \]