What is the general solution for 1=2cos(3θ) ?

The general solution for 1=2cos(3θ) is θ= pi/9+(2pin)/3 ,θ=(5pi)/9+(2pin)/3

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1=2cos(3θ)

1 = 2 cos (3 θ)

θ = \frac{π}{9} + \frac{2 πn}{3}, θ = \frac{5 π}{9} + \frac{2 πn}{3}

Solution steps

1 = 2 cos (3 θ)

Switch sides

2 cos (3 θ) = 1

Divide both sides by

2

cos (3 θ) = \frac{1}{2}

General solutions for

cos (3 θ) = \frac{1}{2}

3 θ = \frac{π}{3} + 2 πn, 3 θ = \frac{5 π}{3} + 2 πn

Solve

3 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{9} + \frac{2 πn}{3}

Solve

3 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{9} + \frac{2 πn}{3}

θ = \frac{π}{9} + \frac{2 πn}{3}, θ = \frac{5 π}{9} + \frac{2 πn}{3}

cos (4 θ) + cos (2 θ) = cos (θ)

(tan (x) + 3) (tan (x) + 2) = 0

2 tan (θ) + 6 = 0

8 = 8 cos (3 θ)

sinh (2 x) = 0

What is the general solution for 1=2cos(3θ) ?
The general solution for 1=2cos(3θ) is θ= pi/9+(2pin)/3 ,θ=(5pi)/9+(2pin)/3