What is the general solution for 8cos(2x)=8-8cos(2x) ?

The general solution for 8cos(2x)=8-8cos(2x) is x= pi/6+pin,x=(5pi)/6+pin

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8cos(2x)=8-8cos(2x)

Solution

8 cos (2 x) = 8 - 8 cos (2 x)

Solution

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

8 cos (2 x) = 8 - 8 cos (2 x)

Solve by substitution

8 cos (2 x) = 8 - 8 cos (2 x)

Let:

cos (2 x) = u

8 u = 8 - 8 u

8 u = 8 - 8 u : u = \frac{1}{2}

8 u = 8 - 8 u

Move

8 u

to the left side

8 u = 8 - 8 u

Add

8 u

to both sides

8 u + 8 u = 8 - 8 u + 8 u

Simplify

16 u = 8

16 u = 8

Divide both sides by

16

16 u = 8

Divide both sides by

16

\frac{16 u}{16} = \frac{8}{16}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = cos (2 x)

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

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

cos (2 x) = \frac{1}{2} : x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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

General solutions for

cos (2 x) = \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}

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

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

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Solve

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8cos(2x)=8-8cos(2x) ?
The general solution for 8cos(2x)=8-8cos(2x) is x= pi/6+pin,x=(5pi)/6+pin