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

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

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2cos(θ)+1=0,0<= θ<= 2pi

2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Degrees

θ = 12 0^{\circ}, θ = 24 0^{\circ}

Solution steps

2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

2 cos (θ) + 1 = 0

Subtract

1

from both sides

2 cos (θ) + 1 - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

t, 4.6 = 0.106 cos (4.36 t)

sin (x) = \frac{23}{24}

5 - 5 cos (x) = 4 sin^{2} (x)

3 cot^{2} (x) - 14 csc (x) - 2 = 0

250 = 100 \cdot 5 \cdot cos (x)

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