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

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

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

Solution

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

Solution

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

Degrees

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

Solution steps

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

Rewrite using trig identities

sin (x) + 2 sin (2 x)

Use the Double Angle identity:

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

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

Simplify

= sin (x) + 4 sin (x) cos (x)

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

Factor

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

sin (x) + 4 cos (x) sin (x)

Factor out common term

sin (x)

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

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

Solving each part separately

sin (x) = 0 or 4 cos (x) + 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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

4 cos (x) + 1 = 0

Move

1

to the right side

4 cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

4 cos (x) = - 1

4 cos (x) = - 1

Divide both sides by

4

4 cos (x) = - 1

Divide both sides by

4

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

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

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

cosh(x)cosh(1)-sinh(x)sinh(1)=2

cosh (x) cosh (1) - sinh (x) sinh (1) = 2

cos^2(x)=(3(1-sin(x)))/2

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2cos(θ)=2cos(3θ)

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cos(t)+cos(2t)=-0.75

cos (t) + cos (2 t) = - 0.75

Frequently Asked Questions (FAQ)

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