What is the general solution for cos(θ)=(-1)/2 ,0<= θ<= 4pi ?

The general solution for cos(θ)=(-1)/2 ,0<= θ<= 4pi is θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(8pi)/3 ,θ=(10pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

cos(θ)=(-1)/2 ,0<= θ<= 4pi

cos (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{8 π}{3}, θ = \frac{10 π}{3}

Degrees

θ = 12 0^{\circ}, θ = 24 0^{\circ}, θ = 48 0^{\circ}, θ = 60 0^{\circ}

Solution steps

cos (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

Apply the fraction rule:

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

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

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

Solutions for the range

0 \leq θ \leq 4 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{8 π}{3}, θ = \frac{10 π}{3}

What is the general solution for cos(θ)=(-1)/2 ,0<= θ<= 4pi ?
The general solution for cos(θ)=(-1)/2 ,0<= θ<= 4pi is θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(8pi)/3 ,θ=(10pi)/3