What is the general solution for 2cos(2θ)+1=0 ?

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

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2cos(2θ)+1=0

2 cos (2 θ) + 1 = 0

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

Solution steps

2 cos (2 θ) + 1 = 0

Move

1

to the right side

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

sec (\frac{5 θ}{4}) = 2

sin^{2} (x) = 1 - cos (x)

3 sec (x) - 4 cos (x) = 0

arctan (x) = \frac{π}{2}

sin^{2} (x) - \frac{1}{2} = 0

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