What is the general solution for -sin(2x)-sin(x)=0,0<= x<2pi ?

Q: What is the general solution for -sin(2x)-sin(x)=0,0<= x<2pi ?

The general solution for -sin(2x)-sin(x)=0,0<= x<2pi is x=0,x=pi,x=(2pi)/3 ,x=(4pi)/3

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Popular Trigonometry >

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

Solution

- sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

Solution

x = 0, x = π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

- sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

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, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

2 cos (x) + 1 = 0, 0 \leq x < 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

2 cos (x) + 1 = 0, 0 \leq x < 2 π

Move

1

to the right side

2 cos (x) + 1 = 0

Subtract

1

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

x = 0, x = π, x = \frac{2 π}{3}, x = \frac{4 π}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -sin(2x)-sin(x)=0,0<= x<2pi ?
The general solution for -sin(2x)-sin(x)=0,0<= x<2pi is x=0,x=pi,x=(2pi)/3 ,x=(4pi)/3