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

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

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-1=2cos(2θ)

- 1 = 2 cos (2 θ)

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

Solution steps

- 1 = 2 cos (2 θ)

Switch sides

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

tan (x) = \frac{10}{12}

0.666 = sin (x)

2 \cdot cos^{2} (x) - sin^{2} (x) + 1 = 0

1 = sinh (x)

15 cos^{2} (θ) + cos (θ) - 2 = 0

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