What is the general solution for 2sqrt(3)+4cos(θ)=0 ?

The general solution for 2sqrt(3)+4cos(θ)=0 is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin

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2sqrt(3)+4cos(θ)=0

23 + 4 cos (θ) = 0

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

Degrees

θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n

Solution steps

23 + 4 cos (θ) = 0

Move

23

to the right side

23 + 4 cos (θ) = 0

Subtract

23

from both sides

23 + 4 cos (θ) - 23 = 0 - 23

Simplify

4 cos (θ) = - 23

4 cos (θ) = - 23

Divide both sides by

4

4 cos (θ) = - 23

Divide both sides by

4

\frac{4 cos ( θ )}{4} = \frac{- 2 3}{4}

Simplify

\frac{4 cos ( θ )}{4} = \frac{- 2 3}{4}

Simplify

\frac{4 cos ( θ )}{4} : cos (θ)

\frac{4 cos ( θ )}{4}

Divide the numbers:

\frac{4}{4} = 1

= cos (θ)

Simplify

\frac{- 2 3}{4} : - \frac{3}{2}

\frac{- 2 3}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 3}{4}

Cancel the common factor:

2

= - \frac{3}{2}

cos (θ) = - \frac{3}{2}

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

2 cos (x) = 1, 0 \leq x < 2 π

cos (x) = - \frac{3}{2}

tan^{2} (x) + 2 tan (x) - 3 = 0

sin (2 x - \frac{π}{3}) = - \frac{1}{2}

2 cos (α) = 3 tan (α)

What is the general solution for 2sqrt(3)+4cos(θ)=0 ?
The general solution for 2sqrt(3)+4cos(θ)=0 is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin