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

The general solution for cos(θ)=-(sqrt(3))/2 ,0<= θ<= 2pi is θ=(5pi)/6 ,θ=(7pi)/6

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cos(θ)=-(sqrt(3))/2 ,0<= θ<= 2pi

cos (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

Degrees

θ = 15 0^{\circ}, θ = 21 0^{\circ}

Solution steps

cos (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

General solutions for

cos (θ) = - \frac{3}{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{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

cos (2 t) = \frac{3}{2}

cos (x) = - 0.3178, 0 \leq x \leq 2 π

arccos (x) = \frac{π}{3}

arccos (x) = \frac{π}{2}

2 cos (2 θ) = 1

What is the general solution for cos(θ)=-(sqrt(3))/2 ,0<= θ<= 2pi ?
The general solution for cos(θ)=-(sqrt(3))/2 ,0<= θ<= 2pi is θ=(5pi)/6 ,θ=(7pi)/6