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

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

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8cos(θ)=-4

8 cos (θ) = - 4

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

8 cos (θ) = - 4

Divide both sides by

8

8 cos (θ) = - 4

Divide both sides by

8

\frac{8 cos ( θ )}{8} = \frac{- 4}{8}

Simplify

\frac{8 cos ( θ )}{8} = \frac{- 4}{8}

Simplify

\frac{8 cos ( θ )}{8} : cos (θ)

\frac{8 cos ( θ )}{8}

Divide the numbers:

\frac{8}{8} = 1

= cos (θ)

Simplify

\frac{- 4}{8} : - \frac{1}{2}

\frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

cos (θ) = - \frac{1}{2}

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

2 sin (3 x + π) = - 1

t, f = cos (a t)

2 cos (x) + 1 = 0, π < x < 4 π

cos (3 x) = 3

4 cos^{2} (θ) - 5 cos (θ) = - 4 cos (θ) + 3

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