What is the general solution for sin(a/2)=(1/2)sin(a) ?

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

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sin(a/2)=(1/2)sin(a)

sin (\frac{a}{2}) = (\frac{1}{2}) sin (a)

a = 4 πn, a = 2 π + 4 πn

Solution steps

sin (\frac{a}{2}) = (\frac{1}{2}) sin (a)

Subtract

\frac{1}{2} sin (a)

from both sides

sin (\frac{a}{2}) - \frac{1}{2} sin (a) = 0

Simplify

sin (\frac{a}{2}) - \frac{1}{2} sin (a) : \frac{2 sin ( \frac{a}{2} ) - sin ( a )}{2}

\frac{2 sin ( \frac{a}{2} ) - sin ( a )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin (\frac{a}{2}) - sin (a) = 0

Let:

u = \frac{a}{2}

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

Rewrite using trig identities

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

Factor

2 sin (u) - 2 cos (u) sin (u) : - 2 sin (u) (cos (u) - 1)

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

Solving each part separately

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

sin (u) = 0 : u = 2 πn, u = π + 2 πn

cos (u) - 1 = 0 : u = 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn

Substitute back

u = \frac{a}{2}

\frac{a}{2} = 2 πn : a = 4 πn

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

a = 4 πn, a = 2 π + 4 πn

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

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

2 + cos^{2} (x) = 5 sin (x)

tan^{3} (3 x) - 2 sin^{3} (3 x) = 0

cot^{5} (x) = \frac{- 1}{( 3 )}

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