What is the general solution for -8cos(θ)=4sqrt(3) ?

The general solution for -8cos(θ)=4sqrt(3) is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin

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-8cos(θ)=4sqrt(3)

- 8 cos (θ) = 43

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

- 8 cos (θ) = 43

Divide both sides by

- 8

- 8 cos (θ) = 43

Divide both sides by

- 8

\frac{- 8 cos ( θ )}{- 8} = \frac{4 3}{- 8}

Simplify

\frac{- 8 cos ( θ )}{- 8} = \frac{4 3}{- 8}

Simplify

\frac{- 8 cos ( θ )}{- 8} : cos (θ)

\frac{- 8 cos ( θ )}{- 8}

Apply the fraction rule:

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

= \frac{8 cos ( θ )}{8}

Divide the numbers:

\frac{8}{8} = 1

= cos (θ)

Simplify

\frac{4 3}{- 8} : - \frac{3}{2}

\frac{4 3}{- 8}

Apply the fraction rule:

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

= - \frac{4 3}{8}

Cancel the common factor:

4

= - \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 sin (3 a) = - 1

sin (b) = 12 (\frac{sin ( 4 2 ^{\circ} )}{22})

arctan (\frac{3}{1}) = tan (θ)

2 sin^{2} (x) - 5 cos (x) + 5 = 0

tan (x) = \frac{20}{40}

What is the general solution for -8cos(θ)=4sqrt(3) ?
The general solution for -8cos(θ)=4sqrt(3) is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin