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

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

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sin(2t)+sin(t)=0

Solution

sin (2 t) + sin (t) = 0

Solution

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

Degrees

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

Solution steps

sin (2 t) + sin (t) = 0

Rewrite using trig identities

sin (2 t) + sin (t)

Use the Double Angle identity:

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

= 2 sin (t) cos (t) + sin (t)

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

Factor

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

sin (t) + 2 cos (t) sin (t)

Factor out common term

sin (t)

= sin (t) (1 + 2 cos (t))

sin (t) (2 cos (t) + 1) = 0

Solving each part separately

sin (t) = 0 or 2 cos (t) + 1 = 0

sin (t) = 0 : t = 2 πn, t = π + 2 πn

sin (t) = 0

General solutions for

sin (t) = 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}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

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

2 cos (t) + 1 = 0

Move

1

to the right side

2 cos (t) + 1 = 0

Subtract

1

from both sides

2 cos (t) + 1 - 1 = 0 - 1

Simplify

2 cos (t) = - 1

2 cos (t) = - 1

Divide both sides by

2

2 cos (t) = - 1

Divide both sides by

2

\frac{2 cos ( t )}{2} = \frac{- 1}{2}

Simplify

cos (t) = - \frac{1}{2}

cos (t) = - \frac{1}{2}

General solutions for

cos (t) = - \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}

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

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

Combine all the solutions

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

Popular Examples

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2 sin^{2} (x) = 3 (cos (x) + 1)

3cot(x/2)=sqrt(3)

3 cot (\frac{x}{2}) = 3

tan^2(θ)-6tan(θ)+5=0

tan^{2} (θ) - 6 tan (θ) + 5 = 0

tan(4θ)=-1

tan (4 θ) = - 1

cot(x)(tan(x)-1)=0

cot (x) (tan (x) - 1) = 0

Frequently Asked Questions (FAQ)

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