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

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

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

Solution

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

Solution

x = 4 πn, x = 2 π + 4 πn, x = π + 4 πn, x = 3 π + 4 πn

Degrees

x = 0^{\circ} + 72 0^{\circ} n, x = 36 0^{\circ} + 72 0^{\circ} n, x = 18 0^{\circ} + 72 0^{\circ} n, x = 54 0^{\circ} + 72 0^{\circ} n

Solution steps

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

Subtract

sin (\frac{x}{2})

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

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

Factor

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

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

Factor out common term

sin (u)

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

Expand

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

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

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

- (1 + cos (2 u))

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

= - 1 - cos (2 u)

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

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

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

Solving each part separately

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

- 1 - (2 cos^{2} (u) - 1) + 2 cos (u) : 2 cos (u) - 2 cos^{2} (u)

- 1 - (2 cos^{2} (u) - 1) + 2 cos (u)

- (2 cos^{2} (u) - 1) : - 2 cos^{2} (u) + 1

- (2 cos^{2} (u) - 1)

Distribute parentheses

= - (2 cos^{2} (u)) - (- 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

- 1 - 2 cos^{2} (u) + 1 + 2 cos (u) : 2 cos (u) - 2 cos^{2} (u)

- 1 - 2 cos^{2} (u) + 1 + 2 cos (u)

Group like terms

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

- 1 + 1 = 0

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

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

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

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

Solve by substitution

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

Let:

cos (u) = u

2 u - 2 u^{2} = 0

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

2 u - 2 u^{2} = 0

Write in the standard form

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

- 2 u^{2} + 2 u = 0

Solve with the quadratic formula

- 2 u^{2} + 2 u = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 2 + 2 = 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{- 2 - 2}{2 ( - 2 )} : 1

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 2 - 2 = - 4

= \frac{- 4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = cos (u)

cos (u) = 0, cos (u) = 1

cos (u) = 0, cos (u) = 1

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

cos (u) = 0

General solutions for

cos (u) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

cos (u) = 1 : u = 2 πn

cos (u) = 1

General solutions for

cos (u) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

u = 0 + 2 πn

u = 0 + 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

Combine all the solutions

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

Combine all the solutions

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

\frac{x}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

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

Simplify

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

x = π + 4 πn

x = π + 4 πn

x = π + 4 πn

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

\frac{x}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

Simplify

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

Multiply fractions:

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

= \frac{3 π 2}{2}

Cancel the common factor:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 4 πn, x = 2 π + 4 πn, x = π + 4 πn, x = 3 π + 4 πn

Since the equation is undefined for:

π + 4 πn, 3 π + 4 πn

x = 4 πn, x = 2 π + 4 πn, x = π + 4 πn, x = 3 π + 4 πn

Popular Examples

sqrt(2)*cos(x/2)-cos(x)=1

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

2sin(θ)cos(θ)+2sin(θ)-2cos(θ)sin(θ)=0

2 sin (θ) cos (θ) + 2 sin (θ) - 2 cos (θ) sin (θ) = 0

2sin(2x)+2cos(x)=0

2 sin (2 x) + 2 cos (x) = 0

cosh(x)= 3/2

cosh (x) = \frac{3}{2}

(1+tan(t))/(sin(t))=0

\frac{1 + tan ( t )}{sin ( t )} = 0

Frequently Asked Questions (FAQ)

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