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

The general solution for 2cos(θ)+1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos(θ)+1=0

2 cos (θ) + 1 = 0

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

Solution steps

2 cos (θ) + 1 = 0

Move

1

to the right side

2 cos (θ) = - 1

Divide both sides by

2

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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

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

sin (16 5^{\circ})

1 - 2 sin (θ) = cos (2 θ)

cos (π)

sec^{2} (1)

What is the general solution for 2cos(θ)+1=0 ?
The general solution for 2cos(θ)+1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin