What is the general solution for sin(2θ)+sin(θ)=0 ?

The general solution for sin(2θ)+sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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Popular >

sin(2θ)+sin(θ)=0

Solution

sin (2 θ) + sin (θ) = 0

Solution

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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (2 θ) + sin (θ) = 0

Rewrite using trig identities

sin (2 θ) + sin (θ)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (θ) cos (θ) + sin (θ)

sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

sin (θ) + 2 cos (θ) sin (θ) : sin (θ) (2 cos (θ) + 1)

sin (θ) + 2 cos (θ) sin (θ)

Factor out common term

sin (θ)

= sin (θ) (1 + 2 cos (θ))

sin (θ) (2 cos (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (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

2 cos (θ) + 1 = 0 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

2 cos (θ) + 1 = 0

Move

1

to the right side

2 cos (θ) + 1 = 0

Subtract

1

from both sides

2 cos (θ) + 1 - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

Simplify

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{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

Combine all the solutions

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

Popular Examples

cos(θ)=(sqrt(3))/2

cos (θ) = \frac{3}{2}

tan(-pi)

tan (- π)

sin(x)=-0.7

sin (x) = - 0.7

cot(30)

cot (3 0^{\circ})

prove (cos(2x))/((cos(x)+sin(x)))=cos(x)-sin(x)prove

\frac{cos ( 2 x )}{( cos ( x ) + sin ( x ) )} = cos (x) - sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)+sin(θ)=0 ?
The general solution for sin(2θ)+sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin