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

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

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2cos(θ/2)=cos(θ)+1

Solution

2 cos (\frac{θ}{2}) = cos (θ) + 1

Solution

θ = π + 4 πn, θ = 3 π + 4 πn, θ = 4 πn

Degrees

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

Solution steps

2 cos (\frac{θ}{2}) = cos (θ) + 1

Subtract

cos (θ) + 1

from both sides

2 cos (\frac{θ}{2}) - cos (θ) - 1 = 0

Let:

u = \frac{θ}{2}

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

Substitute back

u = \frac{θ}{2}

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

\frac{θ}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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

θ = π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

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

\frac{θ}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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

θ = 3 π + 4 πn

θ = 3 π + 4 πn

θ = 3 π + 4 πn

\frac{θ}{2} = 2 πn : θ = 4 πn

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot 2 πn

Simplify

θ = 4 πn

θ = 4 πn

θ = π + 4 πn, θ = 3 π + 4 πn, θ = 4 πn

Popular Examples

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1/(sqrt(3))=tan(x)

\frac{1}{3} = tan (x)

sin(θ)=-(sqrt(5))/5

sin (θ) = - \frac{5}{5}

0.9=cos(x)

0.9 = cos (x)

cos(x)= 8/9

cos (x) = \frac{8}{9}

Frequently Asked Questions (FAQ)

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