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

The general solution for 2cos(3θ)=-1 is θ=(2pi)/9+(2pin)/3 ,θ=(4pi)/9+(2pin)/3

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2cos(3θ)=-1

2 cos (3 θ) = - 1

θ = \frac{2 π}{9} + \frac{2 πn}{3}, θ = \frac{4 π}{9} + \frac{2 πn}{3}

Solution steps

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

Solve

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

Solve

3 θ = \frac{4 π}{3} + 2 πn : θ = \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}, θ = \frac{4 π}{9} + \frac{2 πn}{3}

1 + sin (2 x) = 0

2 cos^{2} (x) + 3 cos (x) - 3 = 0

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

cos (2 x) = \frac{2}{2}, 0 \leq x \leq 2 π

sin (2 x) cos (x) = 0

What is the general solution for 2cos(3θ)=-1 ?
The general solution for 2cos(3θ)=-1 is θ=(2pi)/9+(2pin)/3 ,θ=(4pi)/9+(2pin)/3