Question

X+y=0 sin^2x+sin^2y=1/2

Answer 1

X+y=0×sin^2x+sin^2y=1/2
x+y+0+sin y^2=1/2
x+y+siny^2=1/2
x=1/2-y-siny^2
x=1/2-y-siny^2;y €R

Answer 2

To find the values of x and y that satisfy the given system of equations, we can use the fact that sin^2(x) + sin^2(y) = 1. We are given sin^2(x) + sin^2(y) = 1/2, which means sin^2(x) + sin^2(y) = sin^2(1) (since 1/2 = sin^2(1)).

Now let's solve the system of equations:

1. x + y = 0
2. sin^2(x) + sin^2(y) = sin^2(1)

From equation 1, we can express y in terms of x: y = -x.

Substitute y = -x into equation 2:

sin^2(x) + sin^2(-x) = sin^2(1)

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation:

sin^2(x) + (1 - sin^2(x)) = sin^2(1)

Simplify:

2sin^2(x) - 1 = sin^2(1)

Now, solve for sin^2(x):

2sin^2(x) = sin^2(1) + 1 sin^2(x) = (sin^2(1) + 1) / 2

Since 0 <= sin^2(x) <= 1, sin^2(1) + 1 must be less than or equal to 2.

sin^2(x) = (sin^2(1) + 1) / 2 <= 2 / 2 = 1

So, (sin^2(1) + 1) / 2 = 1, and sin^2(1) + 1 = 2.

Now, plug sin^2(1) + 1 back into the equation:

2sin^2(x) - 1 = sin^2(1) 2sin^2(x) - 1 = 2

Now, solve for sin^2(x):

2sin^2(x) = 2 + 1 2sin^2(x) = 3 sin^2(x) = 3/2

However, there's a problem here because sin^2(x) cannot be greater than 1. This means there are no real solutions for x and y that satisfy the given system of equations. The system appears to be inconsistent. Double-checking the system of equations or confirming the values might be necessary.