Question

Solve the equation for solutions in interval [0 , 2pi)
Sin(x/2)= square root 2 - sin (x/2)

Answer 1

Answer: We can simplify the given equation as follows:

sin(x/2) = sqrt(2) - sin(x/2)

2sin(x/2) = sqrt(2)

sin(x/2) = sqrt(2)/2

x/2 = pi/4 or x/2 = 3pi/4 (using the standard angles for sin)

x = pi/2 or x = 3pi/2

However, we need to check if these solutions satisfy the original equation since we simplified it along the way.

For x = pi/2, we have:

sin(pi/4) = sqrt(2) - sin(pi/4)

sqrt(2)/2 = sqrt(2) - sqrt(2)/2

sqrt(2)/2 = sqrt(2)/2

This is true, so x = pi/2 is a solution.

For x = 3pi/2, we have:

sin(3pi/4) = sqrt(2) - sin(3pi/4)

-sqrt(2)/2 = sqrt(2) - (-sqrt(2)/2)

-sqrt(2)/2 = sqrt(2)/2

This is not true, so x = 3pi/2 is not a solution.

Therefore, the only solution in the interval [0, 2pi) is x = pi/2.

Explanation: