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

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

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1/2+cos(θ)=0

\frac{1}{2} + cos (θ) = 0

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1}{2} + cos (θ) = 0

Move

\frac{1}{2}

to the right side

\frac{1}{2} + cos (θ) = 0

Subtract

\frac{1}{2}

from both sides

\frac{1}{2} + cos (θ) - \frac{1}{2} = 0 - \frac{1}{2}

Simplify

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

(2 cos (x) - 1) (2 sin (x) + 3) = 0

2 sin (2 x) = 2 cos (x)

tan (\frac{1}{2} x) + 1 = 0

2 sin^{2} (θ) + 3 sin (θ) = - 1

tan (4 x) - 3 = 0

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