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

The general solution for sin(t)+2sin(2t)=0 is t=2pin,t=pi+2pin,t=1.82347…+2pin,t=-1.82347…+2pin

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Popular >

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

Solution

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

Solution

t = 2 πn, t = π + 2 πn, t = 1.82347 \dots + 2 πn, t = - 1.82347 \dots + 2 πn

Degrees

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = 104.47751 \dots^{\circ} + 36 0^{\circ} n, t = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

sin (t) + 2 sin (2 t)

Use the Double Angle identity:

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

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

Simplify

= sin (t) + 4 sin (t) cos (t)

sin (t) + 4 cos (t) sin (t) = 0

Factor

sin (t) + 4 cos (t) sin (t) : sin (t) (4 cos (t) + 1)

sin (t) + 4 cos (t) sin (t)

Factor out common term

sin (t)

= sin (t) (1 + 4 cos (t))

sin (t) (4 cos (t) + 1) = 0

Solving each part separately

sin (t) = 0 or 4 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

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

4 cos (t) + 1 = 0

Move

1

to the right side

4 cos (t) + 1 = 0

Subtract

1

from both sides

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

Simplify

4 cos (t) = - 1

4 cos (t) = - 1

Divide both sides by

4

4 cos (t) = - 1

Divide both sides by

4

\frac{4 cos ( t )}{4} = \frac{- 1}{4}

Simplify

cos (t) = - \frac{1}{4}

cos (t) = - \frac{1}{4}

Apply trig inverse properties

cos (t) = - \frac{1}{4}

General solutions for

cos (t) = - \frac{1}{4}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

t = arccos (- \frac{1}{4}) + 2 πn, t = - arccos (- \frac{1}{4}) + 2 πn

t = arccos (- \frac{1}{4}) + 2 πn, t = - arccos (- \frac{1}{4}) + 2 πn

Combine all the solutions

t = 2 πn, t = π + 2 πn, t = arccos (- \frac{1}{4}) + 2 πn, t = - arccos (- \frac{1}{4}) + 2 πn

Show solutions in decimal form

t = 2 πn, t = π + 2 πn, t = 1.82347 \dots + 2 πn, t = - 1.82347 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(t)+2sin(2t)=0 ?
The general solution for sin(t)+2sin(2t)=0 is t=2pin,t=pi+2pin,t=1.82347…+2pin,t=-1.82347…+2pin