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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Subtract

1 - cos (x)

from both sides

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

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

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

Solve by substitution

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

sin (u) = 0 : u = 2 πn, u = π + 2 πn

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

Combine all the solutions

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

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

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

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

cot (2 θ) = \frac{3}{4}

x, 0 = 4 - sec^{2} (x)

tan (x) = - \frac{4}{33}

2 cot^{2} (x) - 11 csc (x) = - 14

5 = tan (x)

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