Question

Solve for θ if 0° ≤ θ < 360°.
sin(theta/2)-cos(theta)=-2

Answer 1

There are three different degrees. One of these is NONE. Thus, leaving only two real solutions.

Answer 2

sin(theta/2) - cos(theta) = -2


sin(x) - cos(2x) = -2 ... Note: I'm letting x = theta/2, so theta = 2x


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


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


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


2*(sin(x))^2 + sin(x) - 1 + 2 = 0


2*(sin(x))^2 + sin(x) + 1 = 0


Now let z = sin(x) to get


2z^2 + z + 1 = 0


Now use the quadratic formula to solve for z


z = (-b+-sqrt(b^2-4ac))/(2a)


z = (-(1)+-sqrt((1)^2-4(2)(1)))/(2(2))


z = (-1+-sqrt(1-(8)))/(4)


z = (-1+-sqrt(-7))/4


z = (-1+sqrt(-7))/4 or z = (-1-sqrt(-7))/4


z = (-1+i*sqrt(7))/4 or z = (-1-i*sqrt(7))/4


Since z is complex, this means that the argument of sine must be complex as well.


So there are no real solutions.

Answer 3

Thank you!!!

Answer 4

You're welcome