SOLUTION: how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula.

Question 609369: how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula.
cos half-angle formula:
cos x/2=√[(1+cosx)/2]
√[(1+cosx)/2]-sinx=0
√[(1+cosx)/2]=sinx
square both sides
(1+cosx)/2=sin^2x=1-cos^2x
1+cosx=2-2cos^2x
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
..
2cosx-1=0
cosx=1/2
x=π/3 and 5π/3
or
cosx+1=0
cosx=-1
x=π
RELATED QUESTIONS
How do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be... (answered by lwsshak3)

How do you solve sin(x/2)-cosx=0 with a restriction of [0,2pi)? We are supposed to be... (answered by lwsshak3,Barbazzo)

Hi, please help me Solve for the angle theta , where 0 is less and equal to theta which... (answered by Alan3354)

Solve the equation in the interval [0, 2pi.) 2 cos2 θ = 2 − cos 2θ I... (answered by stanbon)

Solve cos(x/2)=-sinx on the interval 0 to... (answered by Alan3354)

how do you solve sin(x/2)-cosx=0 with a restriction of... (answered by lwsshak3)

how do you solve sin(x/2)+cosx=0 with a restriction of... (answered by lwsshak3,Barbazzo)

Solve using the fact that tanx = sinx/cosx and the Pythagorean identity on the domain [0, (answered by lwsshak3)

Using the unit circle to solve for x in radians, and withthe restriction of [0,2pi), how... (answered by Alan3354)