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

The general solution for cos(3θ)=1,0<= θ<= 2pi is θ=0,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=2pi

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cos(3θ)=1,0<= θ<= 2pi

cos (3 θ) = 1, 0 \leq θ \leq 2 π

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

Degrees

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

Solution steps

cos (3 θ) = 1, 0 \leq θ \leq 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 = 0 + 2 πn

3 θ = 0 + 2 πn

Solve

3 θ = 0 + 2 πn : θ = \frac{2 πn}{3}

3 θ = 0 + 2 πn

0 + 2 πn = 2 πn

3 θ = 2 πn

Divide both sides by

3

3 θ = 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{2 πn}{3}

Simplify

θ = \frac{2 πn}{3}

θ = \frac{2 πn}{3}

θ = \frac{2 πn}{3}

Solutions for the range

0 \leq θ \leq 2 π

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

3 \cdot sin (\frac{x}{2}) + cos (x) = 1

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

tan^{3} (θ) + tan^{2} (θ) - 3 tan (θ) - 3 = 0

x, sec (x) tan (x) = 0

csc (x) = \frac{1}{4}

What is the general solution for cos(3θ)=1,0<= θ<= 2pi ?
The general solution for cos(3θ)=1,0<= θ<= 2pi is θ=0,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=2pi