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

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

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sin(x)sin(x/2)=1-cos(x)

sin (x) sin (\frac{x}{2}) = 1 - cos (x)

x = 4 πn, x = 2 π + 4 πn

Solution steps

sin (x) sin (\frac{x}{2}) = 1 - cos (x)

Subtract

1 - cos (x)

from both sides

sin (x) sin (\frac{x}{2}) - 1 + cos (x) = 0

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

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

Factor

- 2 sin^{2} (u) + sin (2 u) sin (u) : sin (u) (- 2 sin (u) + sin (2 u))

sin (u) (- 2 sin (u) + sin (2 u)) = 0

Solving each part separately

sin (u) = 0 or - 2 sin (u) + sin (2 u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

- 2 sin (u) + sin (2 u) = 0 : u = 2 πn, u = π + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn

Substitute back

u = \frac{x}{2}

\frac{x}{2} = 2 πn : x = 4 πn

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

x = 4 πn, x = 2 π + 4 πn

24 tan^{2} (x) - 8 = 0

2 sin (x - \frac{π}{3}) = 1

tan (θ) = 2.7

4 (1 + sin (θ)) = cos^{2} (θ)

4 csc (θ) + 4 = 0

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