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

The general solution for cos(θ/2)-cos(θ)=1 is θ=pi+4pin,θ=3pi+4pin,θ=(2pi)/3+4pin,θ=(10pi)/3+4pin

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cos(θ/2)-cos(θ)=1

cos (\frac{θ}{2}) - cos (θ) = 1

θ = π + 4 πn, θ = 3 π + 4 πn, θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

Solution steps

cos (\frac{θ}{2}) - cos (θ) = 1

Subtract

1

from both sides

cos (\frac{θ}{2}) - cos (θ) - 1 = 0

Let:

u = \frac{θ}{2}

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

Rewrite using trig identities

cos (u) - 2 cos^{2} (u) = 0

Solve by substitution

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

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

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

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

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

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

θ = π + 4 πn, θ = 3 π + 4 πn, θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

cot (θ) = \frac{4}{3}

0 = sin (x) - cos (x)

1 - tan^{2} (x) = sec^{2} (x)

cos (2 θ - \frac{π}{2}) = - 1, 0 \leq θ < 2 π

sinh (x) = - \frac{3}{4}

What is the general solution for cos(θ/2)-cos(θ)=1 ?
The general solution for cos(θ/2)-cos(θ)=1 is θ=pi+4pin,θ=3pi+4pin,θ=(2pi)/3+4pin,θ=(10pi)/3+4pin