Find the values of x and y, lying between 0 and 360 which satisfy the equations,
`sin2x^(@)=0.6428`

Answer

To solve the equation \( \sin(2x) = 0.648 \) for \( x \) in the range \( 0 \) to \( 360 \) degrees, we can follow these steps:

### Step 1: Identify the angle whose sine is \( 0.648 \)
We know that:
\[ \sin(40^\circ) \approx 0.648 \]
Thus, we can rewrite the equation as:
\[ \sin(2x) = \sin(40^\circ) \]

### Step 2: Use the sine identity
From the identity \( \sin A = \sin B \), we have:
\[ 2x = n \cdot 180^\circ + (-1)^n \cdot 40^\circ \]
where \( n \) is any integer.

### Step 3: Solve for \( x \)
Now, we can express \( x \) as:
\[ x = \frac{n \cdot 180^\circ + (-1)^n \cdot 40^\circ}{2} \]

### Step 4: Substitute values of \( n \)
We will substitute different integer values of \( n \) to find all possible values of \( x \) within the range \( 0 \) to \( 360 \) degrees.

#### For \( n = 0 \):
\[ x = \frac{0 \cdot 180^\circ + 40^\circ}{2} = \frac{40^\circ}{2} = 20^\circ \]

#### For \( n = 1 \):
\[ x = \frac{1 \cdot 180^\circ - 40^\circ}{2} = \frac{180^\circ - 40^\circ}{2} = \frac{140^\circ}{2} = 70^\circ \]

#### For \( n = 2 \):
\[ x = \frac{2 \cdot 180^\circ + 40^\circ}{2} = \frac{360^\circ + 40^\circ}{2} = \frac{400^\circ}{2} = 200^\circ \]

#### For \( n = 3 \):
\[ x = \frac{3 \cdot 180^\circ - 40^\circ}{2} = \frac{540^\circ - 40^\circ}{2} = \frac{500^\circ}{2} = 250^\circ \]

#### For \( n = 4 \):
\[ x = \frac{4 \cdot 180^\circ + 40^\circ}{2} = \frac{720^\circ + 40^\circ}{2} = \frac{760^\circ}{2} = 380^\circ \]
Since \( 380^\circ \) is outside the range \( 0 \) to \( 360 \), we do not consider this value.

### Step 5: List the valid solutions
The valid solutions for \( x \) in the range \( 0 \) to \( 360 \) degrees are:
\[ x = 20^\circ, 70^\circ, 200^\circ, 250^\circ \]

### Final Answer:
The values of \( x \) that satisfy the equation \( \sin(2x) = 0.648 \) are:
\[ x = 20^\circ, 70^\circ, 200^\circ, 250^\circ \]