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

Q: What is the general solution for sin(2θ)+sin(θ)=0,0<= θ<2pi ?

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

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Popular >

sin(2θ)+sin(θ)=0,0<= θ<2pi

Solution

sin (2 θ) + sin (θ) = 0, 0 \leq θ < 2 π

Solution

θ = 0, θ = π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Degrees

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

Solution steps

sin (2 θ) + sin (θ) = 0, 0 \leq θ < 2 π

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, 0 \leq θ < 2 π : θ = 0, θ = π

sin (θ) = 0, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = 0, θ = π

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

2 cos (θ) + 1 = 0, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

θ = 0, θ = π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

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5sin^2(θ)-11sin(θ)+2=0

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tan^2(θ)+3sec(θ)=-3

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

sin(2x)cos(x)+sin(x)=0

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Frequently Asked Questions (FAQ)

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