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

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

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2cos(2θ)=-1

2 cos (2 θ) = - 1

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

Solution steps

2 cos (2 θ) = - 1

Divide both sides by

2

cos (2 θ) = - \frac{1}{2}

General solutions for

cos (2 θ) = - \frac{1}{2}

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

Solve

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

Solve

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

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

3 cot^{2} (x) = 3 cot (x)

sin (7 x) = \frac{3}{2}

- 1.9 sin (t) = 0.4 cos (2 t)

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

sin (2 t) = 1

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