What is the general solution for 8cos(θ)-4=0 ?

The general solution for 8cos(θ)-4=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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8cos(θ)-4=0

8 cos (θ) - 4 = 0

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

8 cos (θ) - 4 = 0

Move

4

to the right side

8 cos (θ) - 4 = 0

Add

4

to both sides

8 cos (θ) - 4 + 4 = 0 + 4

Simplify

8 cos (θ) = 4

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

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{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

2 (1 + cos (θ)) = 0

0 = cos (π x) \cdot π + sin (π x) π

\frac{sin ( x )}{2} = 1

x, tan (x) = - 3

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

What is the general solution for 8cos(θ)-4=0 ?
The general solution for 8cos(θ)-4=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin