What is the general solution for sin(2x)+sin(x)=0,0<= x<= 360 ?

Q: What is the general solution for sin(2x)+sin(x)=0,0<= x<= 360 ?

The general solution for sin(2x)+sin(x)=0,0<= x<= 360 is x=0,x=180,x=360,x=120,x=240

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sin(2x)+sin(x)=0,0<= x<= 360

Solution

sin (2 x) + sin (x) = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Radians

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

Solution steps

sin (2 x) + sin (x) = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

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 \leq 36 0^{\circ} : x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}

sin (x) = 0, 0 \leq x \leq 36 0^{\circ}

General solutions for

sin (x) = 0

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solve

x = 0 + 36 0^{\circ} n : x = 36 0^{\circ} n

x = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

x = 36 0^{\circ} n

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

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}

2 cos (x) + 1 = 0, 0 \leq x \leq 36 0^{\circ} : x = 12 0^{\circ}, x = 24 0^{\circ}

2 cos (x) + 1 = 0, 0 \leq x \leq 36 0^{\circ}

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

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

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

Combine all the solutions

x = 0, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Popular Examples

cos^2(x)=cos(x)

cos^{2} (x) = cos (x)

cot(x)=-sqrt(3)

cot (x) = - 3

csc((2pi)/3)

csc (\frac{2 π}{3})

cos(x)= 1/4

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

cot((3pi)/4)

cot (\frac{3 π}{4})

Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)+sin(x)=0,0<= x<= 360 ?
The general solution for sin(2x)+sin(x)=0,0<= x<= 360 is x=0,x=180,x=360,x=120,x=240