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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1-cos(θ)= 1/2

Solution

1 - cos (θ) = \frac{1}{2}

Solution

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

Degrees

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

Solution steps

1 - cos (θ) = \frac{1}{2}

Move

1

to the right side

1 - cos (θ) = \frac{1}{2}

Subtract

1

from both sides

1 - cos (θ) - 1 = \frac{1}{2} - 1

Simplify

1 - cos (θ) - 1 = \frac{1}{2} - 1

Simplify

1 - cos (θ) - 1 : - cos (θ)

1 - cos (θ) - 1

Add similar elements:

1 - 1 = 0

= - cos (θ)

Simplify

\frac{1}{2} - 1 : - \frac{1}{2}

\frac{1}{2} - 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- 1 \cdot 2 + 1 = - 1

- 1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

- cos (θ) = - \frac{1}{2}

- cos (θ) = - \frac{1}{2}

- cos (θ) = - \frac{1}{2}

Divide both sides by

- 1

- cos (θ) = - \frac{1}{2}

Divide both sides by

- 1

\frac{- cos ( θ )}{- 1} = \frac{- \frac{1}{2}}{- 1}

Simplify

\frac{- cos ( θ )}{- 1} = \frac{- \frac{1}{2}}{- 1}

Simplify

\frac{- cos ( θ )}{- 1} : cos (θ)

\frac{- cos ( θ )}{- 1}

Apply the fraction rule:

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

= \frac{cos ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos (θ)

Simplify

\frac{- \frac{1}{2}}{- 1} : \frac{1}{2}

\frac{- \frac{1}{2}}{- 1}

Apply the fraction rule:

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

= \frac{\frac{1}{2}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{1}{2}

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

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}

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

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

Popular Examples

(tan(θ)-1)(cos(θ)+1)=0

(tan (θ) - 1) (cos (θ) + 1) = 0

sin^2(θ)=cos^2(θ)

sin^{2} (θ) = cos^{2} (θ)

solvefor x,sin(x)=0 solve for

x, sin (x) = 0

cos(θ)= 2/5

cos (θ) = \frac{2}{5}

2cos^2(x)=cos(x)

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

Frequently Asked Questions (FAQ)

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