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

The general solution for 2sin(θ)=-1,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(θ)=-1,0<= θ<2pi

2 sin (θ) = - 1, 0 \leq θ < 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 21 0^{\circ}, θ = 33 0^{\circ}

Solution steps

2 sin (θ) = - 1, 0 \leq θ < 2 π

Divide both sides by

2

2 sin (θ) = - 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{- 1}{2}

Simplify

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

cos (16 x) = 0

sin (θ) = - \frac{35}{6}

0.6 = cos (x)

3 tan^{3} (x) - 1 = 0

- 2 csc^{2} (2 x) = 0

What is the general solution for 2sin(θ)=-1,0<= θ<2pi ?
The general solution for 2sin(θ)=-1,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6