What is the general solution for -cos(θ)+cos(2θ)=0 ?

The general solution for -cos(θ)+cos(2θ)=0 is θ=(4pin)/3 ,θ=(2pi)/3+(4pin)/3

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-cos(θ)+cos(2θ)=0

Solution

- cos (θ) + cos (2 θ) = 0

Solution

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

Degrees

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

Solution steps

- cos (θ) + cos (2 θ) = 0

Rewrite using trig identities

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

Simplify

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

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

Add similar elements:

2 θ + θ = 3 θ

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

Add similar elements:

2 θ - θ = θ

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

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

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

Solving each part separately

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

sin (\frac{3 θ}{2}) = 0 : θ = \frac{4 πn}{3}, θ = \frac{2 π}{3} + \frac{4 πn}{3}

sin (\frac{3 θ}{2}) = 0

General solutions for

sin (\frac{3 θ}{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 θ}{2} = 0 + 2 πn, \frac{3 θ}{2} = π + 2 πn

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

Solve

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

\frac{3 θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{3 θ}{2} = 2 πn

Multiply both sides by

2

\frac{3 θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 \cdot 3 θ}{2} = 2 \cdot 2 πn

Simplify

3 θ = 4 πn

3 θ = 4 πn

Divide both sides by

3

3 θ = 4 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{4 πn}{3}

Simplify

θ = \frac{4 πn}{3}

θ = \frac{4 πn}{3}

Solve

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

\frac{3 θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{3 θ}{2} = π + 2 πn

Multiply both sides by

2

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

Simplify

3 θ = 2 π + 4 πn

3 θ = 2 π + 4 πn

Divide both sides by

3

3 θ = 2 π + 4 πn

Divide both sides by

3

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

Simplify

θ = \frac{2 π}{3} + \frac{4 πn}{3}

θ = \frac{2 π}{3} + \frac{4 πn}{3}

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

sin (\frac{θ}{2}) = 0 : θ = 4 πn, θ = 2 π + 4 πn

sin (\frac{θ}{2}) = 0

General solutions for

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

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

Solve

\frac{θ}{2} = 0 + 2 πn : θ = 4 πn

\frac{θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot 2 πn

Simplify

θ = 4 πn

θ = 4 πn

Solve

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

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

Simplify

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = 4 πn, θ = 2 π + 4 πn

Combine all the solutions

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

Merge Overlapping Intervals

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

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Frequently Asked Questions (FAQ)

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