What is the general solution for 1-cos(θ)=(2pi)/3 ?

The general solution for 1-cos(θ)=(2pi)/3 is No Solution for θ\in\mathbb{R}

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1-cos(θ)=(2pi)/3

1 - cos (θ) = \frac{2 π}{3}

No Solution for θ \in R

Solution steps

1 - cos (θ) = \frac{2 π}{3}

Move

1

to the right side

1 - cos (θ) = \frac{2 π}{3}

Subtract

1

from both sides

1 - cos (θ) - 1 = \frac{2 π}{3} - 1

Simplify

- cos (θ) = \frac{2 π}{3} - 1

- cos (θ) = \frac{2 π}{3} - 1

Divide both sides by

- 1

- cos (θ) = \frac{2 π}{3} - 1

Divide both sides by

- 1

\frac{- cos ( θ )}{- 1} = \frac{\frac{2 π}{3}}{- 1} - \frac{1}{- 1}

Simplify

\frac{- cos ( θ )}{- 1} = \frac{\frac{2 π}{3}}{- 1} - \frac{1}{- 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{2 π}{3}}{- 1} - \frac{1}{- 1} : - \frac{2 π - 3}{3}

\frac{\frac{2 π}{3}}{- 1} - \frac{1}{- 1}

Apply rule

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

= \frac{\frac{2 π}{3} - 1}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{2 π}{3} - 1}{1}

Join

\frac{2 π}{3} - 1 : \frac{2 π - 3}{3}

\frac{2 π}{3} - 1

Convert element to fraction:

1 = \frac{1 \cdot 3}{3}

= \frac{2 π}{3} - \frac{1 \cdot 3}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π - 1 \cdot 3}{3}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{2 π - 3}{3}

= - \frac{\frac{2 π - 3}{3}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= - \frac{2 π - 3}{3}

cos (θ) = - \frac{2 π - 3}{3}

cos (θ) = - \frac{2 π - 3}{3}

cos (θ) = - \frac{2 π - 3}{3}

- 1 \leq cos (x) \leq 1

No Solution for θ \in R

0.2727 = sec (2.746 x) - 1

23 sin (2 θ) = 3, 0 \leq θ < 2 π

cos (x) - 3 = 3 cos (x)

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

sec (x) = 1.86798

What is the general solution for 1-cos(θ)=(2pi)/3 ?
The general solution for 1-cos(θ)=(2pi)/3 is No Solution for θ\in\mathbb{R}