What is the general solution for sin(θ)(sin(θ)+1)=0 ?

The general solution for sin(θ)(sin(θ)+1)=0 is θ=2pin,θ=pi+2pin,θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)(sin(θ)+1)=0

sin (θ) (sin (θ) + 1) = 0

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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) (sin (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or sin (θ) + 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (θ) + 1 = 0 : θ = \frac{3 π}{2} + 2 πn

sin (θ) + 1 = 0

Move

1

to the right side

sin (θ) + 1 = 0

Subtract

1

from both sides

sin (θ) + 1 - 1 = 0 - 1

Simplify

sin (θ) = - 1

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

6 tan^{2} (x) + 13 tan (x) + 6 = 0

0 = arctan (x) + \frac{π}{2}

2 tan^{2} (θ) = sec^{2} (θ) + 2

5^{s i n (2 x - \frac{x}{4})} = 1

5 + tan (x) = 4

What is the general solution for sin(θ)(sin(θ)+1)=0 ?
The general solution for sin(θ)(sin(θ)+1)=0 is θ=2pin,θ=pi+2pin,θ=(3pi)/2+2pin