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}

Solution steps

2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

2 cos (θ) = - 1

Divide both sides by

2

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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