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

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

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2cos(θ)=-sqrt(3),0<= θ<2pi

2 cos (θ) = - 3, 0 \leq θ < 2 π

θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

Degrees

θ = 15 0^{\circ}, θ = 21 0^{\circ}

Solution steps

2 cos (θ) = - 3, 0 \leq θ < 2 π

Divide both sides by

2

2 cos (θ) = - 3

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 3}{2}

Simplify

cos (θ) = - \frac{3}{2}

cos (θ) = - \frac{3}{2}

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 < 2 π

θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

2 cos (3 θ) = - 3, 0 \leq θ \leq 2 π

csc (θ) = 10

cos (2 x) = \frac{2}{2}, (0, 2 π)

tan (2 x) = 2 tan (x)

- 4 cos (c) - 12 = 2 cos (c) - 6

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