What is the general solution for cos(θ)=-1/2 ,0<= θ<2pi ?

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

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cos(θ)=-1/2 ,0<= θ<2pi

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

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Degrees

θ = 12 0^{\circ}, θ = 24 0^{\circ}

Solution steps

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

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

2 tan (x) - 16 = 0, 0 \leq x < \frac{π}{2}

x, sin (x) = \frac{2}{2}

4 sec (x) + 6 = - 2

tan (2 x) = \frac{1}{4}

tan^{2} (x) = sin^{2} (x) + cos^{2} (x)

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