What is the general solution for 1-cos(θ)=sin(θ/2) ?

The general solution for 1-cos(θ)=sin(θ/2) is θ= pi/3+4pin,θ=(5pi)/3+4pin,θ=4pin,θ=2pi+4pin

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1-cos(θ)=sin(θ/2)

1 - cos (θ) = sin (\frac{θ}{2})

θ = \frac{π}{3} + 4 πn, θ = \frac{5 π}{3} + 4 πn, θ = 4 πn, θ = 2 π + 4 πn

Solution steps

1 - cos (θ) = sin (\frac{θ}{2})

Subtract

sin (\frac{θ}{2})

from both sides

1 - cos (θ) - sin (\frac{θ}{2}) = 0

Let:

u = \frac{θ}{2}

1 - cos (2 u) - sin (u) = 0

Rewrite using trig identities

- sin (u) + 2 sin^{2} (u) = 0

Solve by substitution

sin (u) = \frac{1}{2}, sin (u) = 0

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

sin (u) = 0 : u = 2 πn, u = π + 2 πn

Combine all the solutions

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = 2 πn, u = π + 2 πn

Substitute back

u = \frac{θ}{2}

\frac{θ}{2} = \frac{π}{6} + 2 πn : θ = \frac{π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{3} + 4 πn

\frac{θ}{2} = 2 πn : θ = 4 πn

\frac{θ}{2} = π + 2 πn : θ = 2 π + 4 πn

θ = \frac{π}{3} + 4 πn, θ = \frac{5 π}{3} + 4 πn, θ = 4 πn, θ = 2 π + 4 πn

cos (\frac{θ}{3} + 2 4^{\circ}) = sin (θ)

- 4 + 4 sin (x) = - 2

sin (x + π) + sin (x - π) = 0

cos (y) - sin (y) = 0

36 sin (x) cos (x) = - 93

What is the general solution for 1-cos(θ)=sin(θ/2) ?
The general solution for 1-cos(θ)=sin(θ/2) is θ= pi/3+4pin,θ=(5pi)/3+4pin,θ=4pin,θ=2pi+4pin