What is the general solution for sin(x)=-1/2 sin(2x) ?

The general solution for sin(x)=-1/2 sin(2x) is x=2pin,x=pi+2pin

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sin(x)=-1/2 sin(2x)

Solution

sin (x) = - \frac{1}{2} sin (2 x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

sin (x) = - \frac{1}{2} sin (2 x)

Subtract

- \frac{1}{2} sin (2 x)

from both sides

sin (x) + \frac{1}{2} sin (2 x) = 0

Simplify

sin (x) + \frac{1}{2} sin (2 x) : \frac{2 sin ( x ) + sin ( 2 x )}{2}

sin (x) + \frac{1}{2} sin (2 x)

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

\frac{1}{2} sin (2 x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sin ( 2 x )}{2}

Multiply:

1 \cdot sin (2 x) = sin (2 x)

= \frac{sin ( 2 x )}{2}

= sin (x) + \frac{sin ( 2 x )}{2}

Convert element to fraction:

sin (x) = \frac{sin ( x ) 2}{2}

= \frac{sin ( x ) \cdot 2}{2} + \frac{sin ( 2 x )}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) \cdot 2 + sin ( 2 x )}{2}

\frac{2 sin ( x ) + sin ( 2 x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

sin (2 x) + 2 sin (x)

Use the Double Angle identity:

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

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

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

Factor

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

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

Factor out common term

2 sin (x)

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

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

Solving each part separately

sin (x) = 0 or 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

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

cot(2x)=1.732

cot (2 x) = 1.732

sin(90-a)= 1/(1.63)

sin (9 0^{\circ} - a) = \frac{1}{1.63}

cot^2(x)-cos^2(x)=-1

cot^{2} (x) - cos^{2} (x) = - 1

sin^2(θ)=2*sin(θ)+3

sin^{2} (θ) = 2 \cdot sin (θ) + 3

tan(3x)=1,-2pi<= x<= 2pi

tan (3 x) = 1, - 2 π \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)=-1/2 sin(2x) ?
The general solution for sin(x)=-1/2 sin(2x) is x=2pin,x=pi+2pin