What is the general solution for solvefor x,sin(2x)=sin(x) ?

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

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Popular >

solvefor x,sin(2x)=sin(x)

Solution

solve for

x, sin (2 x) = sin (x)

Solution

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

Degrees

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

Solution steps

sin (2 x) = sin (x)

Subtract

sin (x)

from both sides

sin (2 x) - sin (x) = 0

Rewrite using trig identities

sin (2 x) - sin (x)

Use the Double Angle identity:

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

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

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

Factor

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

- sin (x) + 2 cos (x) sin (x)

Factor out common term

sin (x)

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

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

Solving each part separately

sin (x) = 0 or 2 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

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

2 cos (x) - 1 = 0

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (x) = \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}

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

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

Combine all the solutions

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

Popular Examples

-3cos(2θ)+7cos(θ)-10=-5

- 3 cos (2 θ) + 7 cos (θ) - 10 = - 5

3cos(2θ)=-9cos(θ)-6

3 cos (2 θ) = - 9 cos (θ) - 6

5sin(x)cos(x)+cos(x)=0

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

1+cos(θ)=2sin^2(θ)

1 + cos (θ) = 2 sin^{2} (θ)

cos(2a)=-4/7

cos (2 a) = - \frac{4}{7}

Frequently Asked Questions (FAQ)

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