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

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

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solvefor θ,cos(θ)=-1/2

solve for

θ, cos (θ) = - \frac{1}{2}

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

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

- 3 sin (b) - 2 = sin (b) - 2

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

tan (2 θ - 1 0^{\circ}) = cot (θ)

sinh (x) = 4

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, 0 \leq θ < 2 π

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