What is the general solution for cos(3θ)=cos(θ) ?

The general solution for cos(3θ)=cos(θ) is θ=pin,θ= pi/2+pin

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cos(3θ)=cos(θ)

Solution

cos (3 θ) = cos (θ)

Solution

θ = πn, θ = \frac{π}{2} + πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos (3 θ) = cos (θ)

Subtract

cos (θ)

from both sides

cos (3 θ) - cos (θ) = 0

Rewrite using trig identities

cos (3 θ) - cos (θ)

Use the Sum to Product identity:

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

= - 2 sin (\frac{3 θ + θ}{2}) sin (\frac{3 θ - θ}{2})

Simplify

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

- 2 sin (\frac{3 θ + θ}{2}) sin (\frac{3 θ - θ}{2})

\frac{3 θ + θ}{2} = 2 θ

\frac{3 θ + θ}{2}

Add similar elements:

3 θ + θ = 4 θ

= \frac{4 θ}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 θ

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

\frac{3 θ - θ}{2} = θ

\frac{3 θ - θ}{2}

Add similar elements:

3 θ - θ = 2 θ

= \frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

= - 2 sin (2 θ) sin (θ)

= - 2 sin (2 θ) sin (θ)

- 2 sin (2 θ) sin (θ) = 0

Solving each part separately

sin (2 θ) = 0 or sin (θ) = 0

sin (2 θ) = 0 : θ = πn, θ = \frac{π}{2} + πn

sin (2 θ) = 0

General solutions for

sin (2 θ) = 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}

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

Divide both sides by

2

2 θ = 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2}

Simplify

θ = πn

θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = πn, θ = \frac{π}{2} + πn, θ = 2 πn, θ = π + 2 πn

Merge Overlapping Intervals

θ = πn, θ = \frac{π}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(3θ)=cos(θ) ?
The general solution for cos(3θ)=cos(θ) is θ=pin,θ= pi/2+pin