What is the general solution for sin(2A)=sin(A) ?

The general solution for sin(2A)=sin(A) is A=2pin,A=pi+2pin,A= pi/3+2pin,A=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2A)=sin(A)

Solution

sin (2 A) = sin (A)

Solution

A = 2 πn, A = π + 2 πn, A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

Degrees

A = 0^{\circ} + 36 0^{\circ} n, A = 18 0^{\circ} + 36 0^{\circ} n, A = 6 0^{\circ} + 36 0^{\circ} n, A = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (2 A) = sin (A)

Subtract

sin (A)

from both sides

sin (2 A) - sin (A) = 0

Rewrite using trig identities

sin (2 A) - sin (A)

Use the Double Angle identity:

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

= 2 sin (A) cos (A) - sin (A)

- sin (A) + 2 cos (A) sin (A) = 0

Factor

- sin (A) + 2 cos (A) sin (A) : sin (A) (2 cos (A) - 1)

- sin (A) + 2 cos (A) sin (A)

Factor out common term

sin (A)

= sin (A) (- 1 + 2 cos (A))

sin (A) (2 cos (A) - 1) = 0

Solving each part separately

sin (A) = 0 or 2 cos (A) - 1 = 0

sin (A) = 0 : A = 2 πn, A = π + 2 πn

sin (A) = 0

General solutions for

sin (A) = 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}

A = 0 + 2 πn, A = π + 2 πn

A = 0 + 2 πn, A = π + 2 πn

Solve

A = 0 + 2 πn : A = 2 πn

A = 0 + 2 πn

0 + 2 πn = 2 πn

A = 2 πn

A = 2 πn, A = π + 2 πn

2 cos (A) - 1 = 0 : A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

2 cos (A) - 1 = 0

Move

1

to the right side

2 cos (A) - 1 = 0

Add

1

to both sides

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

Simplify

2 cos (A) = 1

2 cos (A) = 1

Divide both sides by

2

2 cos (A) = 1

Divide both sides by

2

\frac{2 cos ( A )}{2} = \frac{1}{2}

Simplify

cos (A) = \frac{1}{2}

cos (A) = \frac{1}{2}

General solutions for

cos (A) = \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}

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

Combine all the solutions

A = 2 πn, A = π + 2 πn, A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

Popular Examples

csc(θ)= 3/2

csc (θ) = \frac{3}{2}

(cos(θ))/(sin(θ))=-1

\frac{cos ( θ )}{sin ( θ )} = - 1

cos(B)= 12/13

cos (B) = \frac{12}{13}

sin(θ)=sqrt(2/9),θ,3r(θ,d)d,tan(2θ)

sin (θ) = \frac{2}{9}, θ, 3 r (θ, d) d, tan (2 θ)

sin(x/2)=0,0<= x<= 2pi

sin (\frac{x}{2}) = 0, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for sin(2A)=sin(A) ?
The general solution for sin(2A)=sin(A) is A=2pin,A=pi+2pin,A= pi/3+2pin,A=(5pi)/3+2pin