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

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

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

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

Solution

cos (2 x) - cos (x) = 0, 0 \leq x \leq 2 π

Solution

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

Degrees

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

Solution steps

cos (2 x) - cos (x) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

cos (2 x) - cos (x)

Use the Sum to Product identity:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

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

Simplify

- 2 sin (\frac{2 x + x}{2}) sin (\frac{2 x - x}{2}) : - 2 sin (\frac{3 x}{2}) sin (\frac{x}{2})

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

Add similar elements:

2 x + x = 3 x

= - 2 sin (\frac{3 x}{2}) sin (\frac{2 x - x}{2})

Add similar elements:

2 x - x = x

= - 2 sin (\frac{3 x}{2}) sin (\frac{x}{2})

= - 2 sin (\frac{3 x}{2}) sin (\frac{x}{2})

- 2 sin (\frac{3 x}{2}) sin (\frac{x}{2}) = 0

Solving each part separately

sin (\frac{3 x}{2}) = 0 or sin (\frac{x}{2}) = 0

sin (\frac{3 x}{2}) = 0, 0 \leq x \leq 2 π : x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

sin (\frac{3 x}{2}) = 0, 0 \leq x \leq 2 π

General solutions for

sin (\frac{3 x}{2}) = 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}

\frac{3 x}{2} = 0 + 2 πn, \frac{3 x}{2} = π + 2 πn

\frac{3 x}{2} = 0 + 2 πn, \frac{3 x}{2} = π + 2 πn

Solve

\frac{3 x}{2} = 0 + 2 πn : x = \frac{4 πn}{3}

\frac{3 x}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{3 x}{2} = 2 πn

Multiply both sides by

2

\frac{3 x}{2} = 2 πn

Multiply both sides by

2

\frac{2 \cdot 3 x}{2} = 2 \cdot 2 πn

Simplify

3 x = 4 πn

3 x = 4 πn

Divide both sides by

3

3 x = 4 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{4 πn}{3}

Simplify

x = \frac{4 πn}{3}

x = \frac{4 πn}{3}

Solve

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

\frac{3 x}{2} = π + 2 πn

Multiply both sides by

2

\frac{3 x}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 \cdot 3 x}{2} = 2 π + 2 \cdot 2 πn

Simplify

3 x = 2 π + 4 πn

3 x = 2 π + 4 πn

Divide both sides by

3

3 x = 2 π + 4 πn

Divide both sides by

3

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

Simplify

x = \frac{2 π}{3} + \frac{4 πn}{3}

x = \frac{2 π}{3} + \frac{4 πn}{3}

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

Solutions for the range

0 \leq x \leq 2 π

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

sin (\frac{x}{2}) = 0, 0 \leq x \leq 2 π : x = 0, x = 2 π

sin (\frac{x}{2}) = 0, 0 \leq x \leq 2 π

General solutions for

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

\frac{x}{2} = 0 + 2 πn, \frac{x}{2} = π + 2 πn

\frac{x}{2} = 0 + 2 πn, \frac{x}{2} = π + 2 πn

Solve

\frac{x}{2} = 0 + 2 πn : x = 4 πn

\frac{x}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{x}{2} = 2 πn

Multiply both sides by

2

\frac{x}{2} = 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot 2 πn

Simplify

x = 4 πn

x = 4 πn

Solve

\frac{x}{2} = π + 2 πn : x = 2 π + 4 πn

\frac{x}{2} = π + 2 πn

Multiply both sides by

2

\frac{x}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

Simplify

x = 2 π + 4 πn

x = 2 π + 4 πn

x = 4 πn, x = 2 π + 4 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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