What is the general solution for cos(3θ)=-1,0<= θ<2pi ?

The general solution for cos(3θ)=-1,0<= θ<2pi is θ= pi/3 ,θ=pi,θ=(5pi)/3

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cos(3θ)=-1,0<= θ<2pi

cos (3 θ) = - 1, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = π, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 18 0^{\circ}, θ = 30 0^{\circ}

Solution steps

cos (3 θ) = - 1, 0 \leq θ < 2 π

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}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = π, θ = \frac{5 π}{3}

2 cos (x) + 3 = 0

cos (θ) = - \frac{4}{9}

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

2 cos^{2} (x) - 4 cos (x) + 2 = 0

cos^{2} (x) - sin^{2} (x) = 1 - 2 sin (x)

What is the general solution for cos(3θ)=-1,0<= θ<2pi ?
The general solution for cos(3θ)=-1,0<= θ<2pi is θ= pi/3 ,θ=pi,θ=(5pi)/3