What is the general solution for cos(3θ)=-1 ?

The general solution for cos(3θ)=-1 is θ= pi/3+(2pin)/3

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cos(3θ)=-1

cos (3 θ) = - 1

θ = \frac{π}{3} + \frac{2 πn}{3}

Degrees

θ = 6 0^{\circ} + 12 0^{\circ} n

Solution steps

cos (3 θ) = - 1

General solutions for

cos (3 θ) = - 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}

3 θ = π + 2 πn

3 θ = π + 2 πn

Solve

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

3 θ = π + 2 πn

Divide both sides by

3

3 θ = π + 2 πn

Divide both sides by

3

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

Simplify

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{π}{3} + \frac{2 πn}{3}

sin (θ) + 1 = 0

sin (- \frac{π}{2})

cot (arcsin (\frac{2}{2}))

sin (\frac{3}{4})

arccot (- 1)

What is the general solution for cos(3θ)=-1 ?
The general solution for cos(3θ)=-1 is θ= pi/3+(2pin)/3