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

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

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 72 0^{\circ} n, x = 36 0^{\circ} + 72 0^{\circ} n, x = 6 0^{\circ} + 72 0^{\circ} n, x = 30 0^{\circ} + 72 0^{\circ} n

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

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

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

= - 1 + 1 - 2 sin^{2} (u) + sin (u)

Simplify

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

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

Solve by substitution

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

Let:

sin (u) = u

u - 2 u^{2} = 0

u - 2 u^{2} = 0 : u = 0, u = \frac{1}{2}

u - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + u = 0

Solve with the quadratic formula

- 2 u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 0 = 1

1^{2} - 4 (- 2) \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 ( - 2 )}, u_{2} = \frac{- 1 - 1}{2 ( - 2 )}

u = \frac{- 1 + 1}{2 ( - 2 )} : 0

\frac{- 1 + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 1 - 1}{2 ( - 2 )} : \frac{1}{2}

\frac{- 1 - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 1}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 0, u = \frac{1}{2}

Substitute back

u = sin (u)

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

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

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

General solutions for

sin (u) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

Solve

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

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

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

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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} = 2 πn

Multiply both sides by

2

\frac{x}{2} = 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot 2 πn

Simplify

x = 4 πn

x = 4 πn

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

\frac{x}{2} = π + 2 πn

Multiply both sides by

2

\frac{x}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

Simplify

x = 2 π + 4 πn

x = 2 π + 4 πn

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

\frac{x}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

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

\frac{x}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{6}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{6}

Cancel the common factor:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

x = \frac{5 π}{3} + 4 πn

x = \frac{5 π}{3} + 4 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

Popular Examples

cot(2θ)= 3/4

cot (2 θ) = \frac{3}{4}

solvefor x,0=4-sec^2(x)solve for

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

tan(x)=-4/(sqrt(33))

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

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

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

5=tan(x)

5 = tan (x)

Frequently Asked Questions (FAQ)

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