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

The general solution for 2sin(2x)-2sin(x)=0 is x=2pin,x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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Popular >

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

Solution

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

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

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

Rewrite using trig identities

2 sin (2 x) - 2 sin (x)

Use the Double Angle identity:

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

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

Simplify

= 4 sin (x) cos (x) - 2 sin (x)

- 2 sin (x) + 4 cos (x) sin (x) = 0

Factor

- 2 sin (x) + 4 cos (x) sin (x) : 2 sin (x) (2 cos (x) - 1)

- 2 sin (x) + 4 cos (x) sin (x)

Rewrite

4

2 \cdot 2

= - 2 sin (x) + 2 \cdot 2 sin (x) cos (x)

Factor out common term

2 sin (x)

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

2 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

(sin(25))/5 =(sin(b))/8

\frac{sin ( 2 5 ^{\circ} )}{5} = \frac{sin ( b )}{8}

cos(x)-cos(3x)=-sqrt(2)sin(x)

cos (x) - cos (3 x) = - 2 sin (x)

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

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

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

sin^{2} (x) + cos^{2} (x) = tan (x)

3tan(3x)=tan(x)

3 tan (3 x) = tan (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2x)-2sin(x)=0 ?
The general solution for 2sin(2x)-2sin(x)=0 is x=2pin,x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin