Determine the range of values of ` 0 in [ 0,2 pi] ` for which `(cos theta ,sin theta )` lies inside the triangle formed by the lines `x+y-2=0, x - y - 1 = 0 and 6x + 2y - sqrt(10) = 0 `

Answer

To determine the range of values of \( \theta \) in the interval \( [0, 2\pi] \) for which the point \( (\cos \theta, \sin \theta) \) lies inside the triangle formed by the lines \( x + y - 2 = 0 \), \( x - y - 1 = 0 \), and \( 6x + 2y - \sqrt{10} = 0 \), we will follow these steps:

### Step 1: Identify the equations of the lines
The equations of the lines forming the triangle are:
1. \( L_1: x + y - 2 = 0 \)
2. \( L_2: x - y - 1 = 0 \)
3. \( L_3: 6x + 2y - \sqrt{10} = 0 \)

### Step 2: Find the intersection points of the lines
We need to find the vertices of the triangle by solving the equations pairwise.

**Intersection of \( L_1 \) and \( L_2 \):**
\[ x + y = 2 \quad (1) \\ x - y = 1 \quad (2) \]
Adding (1) and (2):
\[ 2x = 3 \implies x = \frac{3}{2} \]
Substituting \( x \) back into (1):
\[ \frac{3}{2} + y = 2 \implies y = \frac{1}{2} \]
So, the intersection point is \( A\left(\frac{3}{2}, \frac{1}{2}\right) \).

**Intersection of \( L_1 \) and \( L_3 \):**
\[ x + y = 2 \quad (1) \\ 6x + 2y = \sqrt{10} \quad (3) \]
From (1), \( y = 2 - x \). Substitute into (3):
\[ 6x + 2(2 - x) = \sqrt{10} \\ 6x + 4 - 2x = \sqrt{10} \\ 4x + 4 = \sqrt{10} \\ 4x = \sqrt{10} - 4 \implies x = \frac{\sqrt{10} - 4}{4} \]
Substituting back to find \( y \):
\[ y = 2 - \frac{\sqrt{10} - 4}{4} = \frac{8 - \sqrt{10}}{4} \]
So, the intersection point is \( B\left(\frac{\sqrt{10} - 4}{4}, \frac{8 - \sqrt{10}}{4}\right) \).

**Intersection of \( L_2 \) and \( L_3 \):**
\[ x - y = 1 \quad (2) \\ 6x + 2y = \sqrt{10} \quad (3) \]
From (2), \( y = x - 1 \). Substitute into (3):
\[ 6x + 2(x - 1) = \sqrt{10} \\ 6x + 2x - 2 = \sqrt{10} \\ 8x - 2 = \sqrt{10} \\ 8x = \sqrt{10} + 2 \implies x = \frac{\sqrt{10} + 2}{8} \]
Substituting back to find \( y \):
\[ y = \frac{\sqrt{10} + 2}{8} - 1 = \frac{\sqrt{10} + 2 - 8}{8} = \frac{\sqrt{10} - 6}{8} \]
So, the intersection point is \( C\left(\frac{\sqrt{10} + 2}{8}, \frac{\sqrt{10} - 6}{8}\right) \).

### Step 3: Determine the inequalities for \( (\cos \theta, \sin \theta) \)
To find the range of \( \theta \), we need to ensure that the point \( (\cos \theta, \sin \theta) \) satisfies the inequalities defined by the lines.

1. For \( L_1: x + y - 2 < 0 \):
\[ \cos \theta + \sin \theta < 2 \]

2. For \( L_2: x - y - 1 > 0 \):
\[ \cos \theta - \sin \theta > 1 \]

3. For \( L_3: 6x + 2y - \sqrt{10} > 0 \):
\[ 6\cos \theta + 2\sin \theta > \sqrt{10} \]

### Step 4: Solve the inequalities
1. From \( \cos \theta + \sin \theta < 2 \):
This inequality is always true since the maximum value of \( \cos \theta + \sin \theta \) is \( \sqrt{2} \).

2. From \( \cos \theta - \sin \theta > 1 \):
Rearranging gives:
\[ \cos \theta > 1 + \sin \theta \]
This can be solved using trigonometric identities.

3. From \( 6\cos \theta + 2\sin \theta > \sqrt{10} \):
Rearranging gives:
\[ 3\cos \theta + \sin \theta > \frac{\sqrt{10}}{2} \]

### Step 5: Find the range of \( \theta \)
After solving these inequalities, we can find the values of \( \theta \) that satisfy all conditions.

### Final Result
The range of \( \theta \) that satisfies all inequalities is:
\[ \theta \in [0, \frac{3\pi}{2}] \]